{"id":2421,"date":"2026-11-13T17:56:43","date_gmt":"2026-11-13T12:26:43","guid":{"rendered":"https:\/\/xamcontent.com\/cisce\/?p=2421"},"modified":"2026-11-13T17:56:44","modified_gmt":"2026-11-13T12:26:44","slug":"icse-class-10-mathematics-2022-semester-1-question-paper-with-solutions","status":"publish","type":"post","link":"https:\/\/xamcontent.com\/cisce\/icse-class-10-mathematics-2022-semester-1-question-paper-with-solutions\/","title":{"rendered":"ICSE Class 10 Mathematics 2022 Semester 1 Question Paper with Solutions"},"content":{"rendered":"Reading Time: <\/span> 22<\/span> minutes<\/span><\/span>\n
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ICSE Class 10 Mathematics PYQ<\/div>\n

Practice the complete ICSE Class 10 Mathematics 2022 Semester 1 Question Paper with Solutions<\/strong>. This 40-mark objective paper includes algebra, commercial mathematics, matrices, arithmetic progression, similarity, recurring deposits and linear inequations with expandable explanations for focused board-exam revision.<\/p>\n<\/div>\n

\n
Year<\/div>
2022 Semester 1<\/div><\/div>\n
Maximum Marks<\/div>
40<\/div><\/div>\n
Time Allowed<\/div>
One hour and a half<\/div><\/div>\n<\/div>\n
\n
\ud83d\udcc4<\/div>
Paper Instructions<\/div>
ICSE Class 10 Mathematics 2022 Semester 1<\/div><\/div><\/div>\n

All questions are compulsory.<\/p>

The marks intended for questions are given in brackets [ ].<\/p>

Select the correct option for each of the following questions.<\/p><\/div>\n<\/div>\n

\n
\u2713<\/div>\n
Solved Objective Paper<\/div>

ICSE Mathematics 2022 Semester 1 Questions with Solutions<\/h2>
40 marks \u2022 All questions are compulsory.<\/div><\/div>\n
40<\/div>
Marks<\/div><\/div>\n<\/div>\n
\n
1<\/div>
Question 1<\/div>
ICSE Class 10 Mathematics 2022 Semester 1<\/div><\/div><\/div>\n
\n
1<\/div>

If \\((\\boldsymbol{x}+\\mathbf{2})\\) is a factor of the polynomial \\(\\boldsymbol{x}^{\\mathbf{3}}-\\boldsymbol{k} \\boldsymbol{x}^{\\mathbf{2}}-\\mathbf{5} \\boldsymbol{x}\\) + 6 then the value of \\(\\boldsymbol{k}\\) is [1]<\/strong><\/p><\/div><\/div>\n

\n
A<\/span>1<\/span><\/div>\n
B<\/span>2<\/span><\/div>\n
C<\/span>3<\/span><\/div>\n
D<\/span>-2<\/span><\/div>\n<\/div>\n
\u25bc<\/span> View Answer & Solution<\/div>\n
Answer & Solution<\/div>

Ans. Option (b) is correct.<\/p>\n

Explanation:<\/strong> <\/p>\n

\\[\n\\begin{aligned}\nx+2 & =0 \\\\\nx & =-2 \\\\\np(x)=x^{3}-k x^{2}-5 x+6 & =0\n\\end{aligned}\n\\]<\/div>\n

Remainder \\(=p(-2)=0\\)<\/p>\n

\\[\n\\begin{aligned}\n(-2)^{3}-k(-2)^{2}-5(-2)+6 & =0 \\\\\n-8-4 k+10+6 & =0 \\\\\n-4 k & =-8 \\\\\nk & =2\n\\end{aligned}\n\\]<\/div><\/div>\n<\/div>\n<\/div>\n
\n
2<\/div>
Question 2<\/div>
ICSE Class 10 Mathematics 2022 Semester 1<\/div><\/div><\/div>\n
\n
2<\/div>

The solution set of the inequation \\(\\boldsymbol{x}-\\mathbf{3} \\geq-\\mathbf{5}, \\boldsymbol{x} \\in \\boldsymbol{R}\\) is [1]<\/strong><\/p><\/div><\/div>\n

\n
A<\/span>\\(\\{x: x>-2, x \\in R\\}\\)<\/span><\/div>\n
B<\/span>\\(\\{x: x \\leq-2, \\mathrm{x} \\in R\\}\\)<\/span><\/div>\n
C<\/span>\\(\\{x: x \\geq-2, x \\in R\\}\\)<\/span><\/div>\n
D<\/span>\\{-2, -1, 0, 1, 2\\}<\/span><\/div>\n<\/div>\n
\u25bc<\/span> View Answer & Solution<\/div>\n
Answer & Solution<\/div>

Ans. Option (c) is correct.<\/p>\n

Explanation:<\/strong> \\(x-3 \\geq-5\\)<\/p>\n

\\[\n\\begin{aligned}\n& x \\geq-5+3 \\\\\n& x \\geq-2\n\\end{aligned}\n\\]<\/div><\/div>\n<\/div>\n<\/div>\n
\n
3<\/div>
Question 3<\/div>
ICSE Class 10 Mathematics 2022 Semester 1<\/div><\/div><\/div>\n
\n
3<\/div>

The product \\(\\boldsymbol{A} \\boldsymbol{B}\\) of two matrices \\(\\boldsymbol{A}\\) and \\(\\boldsymbol{B}\\) is possible if [1]<\/strong><\/p><\/div><\/div>\n

\n
A<\/span>\\(A\\) and \\(B\\) have the same number of rows.<\/span><\/div>\n
B<\/span>the number of columns of \\(A\\) is equal to the number of rows of \\(B\\).<\/span><\/div>\n
C<\/span>the number of rows of \\(A\\) is equal to the number of columns of \\(B\\).<\/span><\/div>\n
D<\/span>\\(A\\) and \\(B\\) have the same number of columns.<\/span><\/div>\n<\/div>\n
\u25bc<\/span> View Answer & Solution<\/div>\n
Answer & Solution<\/div>

Ans. Option (b) is correct.<\/p>\n

Explanation:<\/strong> The product of \\(A B\\) of two matrices \\(A\\) and \\(B\\) is possible if the number of column in \\(A\\) is equal to the number of rows in \\(B\\).<\/p><\/div>\n<\/div>\n<\/div>\n

\n
4<\/div>
Question 4<\/div>
ICSE Class 10 Mathematics 2022 Semester 1<\/div><\/div><\/div>\n
\n
4<\/div>

If 70, 75, 80, 85 are the first four terms of an Arithmetic Progression, then the \\(\\mathbf{1 0}^{\\text {th }}\\) term is [1]<\/strong><\/p><\/div><\/div>\n

\n
A<\/span>35<\/span><\/div>\n
B<\/span>25<\/span><\/div>\n
C<\/span>115<\/span><\/div>\n
D<\/span>105<\/span><\/div>\n<\/div>\n
\u25bc<\/span> View Answer & Solution<\/div>\n
Answer & Solution<\/div>

Ans. Option (c) is correct.<\/p>\n

Explanation:<\/strong> \\(a=70, d=75-70=5\\)<\/p>\n

\\[\n\\begin{aligned}\na_{n} & =a+(n-1) d \\\\\na_{10} & =70+(10-1) 5 \\\\\n& =70+45\n\\end{aligned}\n\\]<\/div>\n
\\[\na_{10}=115\n\\]<\/div><\/div>\n<\/div>\n<\/div>\n
\n
5<\/div>
Question 5<\/div>
ICSE Class 10 Mathematics 2022 Semester 1<\/div><\/div><\/div>\n
\n
5<\/div>

The selling price of a shirt excluding GST is \u20b9 800. If the rate of GST is 12% then the total price of the shirt is [1]<\/strong><\/p><\/div><\/div>\n

\n
A<\/span>\u20b9704<\/span><\/div>\n
B<\/span>\u20b996<\/span><\/div>\n
C<\/span>\u20b9896<\/span><\/div>\n
D<\/span>\u20b9848<\/span><\/div>\n<\/div>\n
\u25bc<\/span> View Answer & Solution<\/div>\n
Answer & Solution<\/div>

Ans. Option (c) is correct.<\/p>\n

Explanation:<\/strong> \\(\\mathrm{SP}=\\text{\u20b9} 800, \\mathrm{GST}=12 \\%\\)<\/p>\n

\\[\n\\begin{aligned}\n\\text { GST } & =\\text{\u20b9} \\frac{12}{100} \\times 800=\\text{\u20b9} 96 \\\\\n\\text { Total Cost } & =800+96=\\text{\u20b9} 896\n\\end{aligned}\n\\]<\/div><\/div>\n<\/div>\n<\/div>\n
\n
6<\/div>
Question 6<\/div>
ICSE Class 10 Mathematics 2022 Semester 1<\/div><\/div><\/div>\n
\n
6<\/div>

Which of the following quadratic equations has 2 and 3 as its roots? [1]<\/strong><\/p><\/div><\/div>\n

\n
A<\/span>\\(x^{2}-5 x+6=0\\)<\/span><\/div>\n
B<\/span>\\(x^{2}+5 x+6=0\\)<\/span><\/div>\n
C<\/span>\\(x^{2}-5 x-6=0\\)<\/span><\/div>\n
D<\/span>\\(x^{2}+5 x-6=0\\)<\/span><\/div>\n<\/div>\n
\u25bc<\/span> View Answer & Solution<\/div>\n
Answer & Solution<\/div>

Ans. Option (a) is correct.<\/p>\n

Explanation:<\/strong> \\(\\because \\quad x^{2}-(\\) sum of roots \\() x-(\\) product of roots \\()=0\\)<\/p>\n

\\[\n\\begin{array}{lr}\n\\therefore & x^{2}-(2+3) x-(2 \\times 3)=0 \\\\\n\\Rightarrow & x^{2}-5 x+6=0\n\\end{array}\n\\]<\/div><\/div>\n<\/div>\n<\/div>\n
\n
7<\/div>
Question 7<\/div>
ICSE Class 10 Mathematics 2022 Semester 1<\/div><\/div><\/div>\n
\n
7<\/div>

If \\(\\boldsymbol{x}\\), 5.4, 5, 9 are in proportion then \\(\\boldsymbol{x}\\) is [1]<\/strong><\/p><\/div><\/div>\n

\n
A<\/span>3<\/span><\/div>\n
B<\/span>9.72<\/span><\/div>\n
C<\/span>25<\/span><\/div>\n
D<\/span>25\/3<\/span><\/div>\n<\/div>\n
\u25bc<\/span> View Answer & Solution<\/div>\n
Answer & Solution<\/div>

Ans. Option (a) is correct.<\/p>\n

Explanation:<\/strong> Numbers are in proportion (given)<\/p>\n

\\[\n\\begin{aligned}\n\\Rightarrow & & \\frac{x}{5.4} & =\\frac{5}{9} \\\\\n\\Rightarrow & & 9 \\times x & =5.4 \\times 5 \\\\\n\\Rightarrow & & x & =\\frac{5.4 \\times 5}{9}=0.6 \\times 5 \\\\\n\\Rightarrow & & x & =3\n\\end{aligned}\n\\]<\/div><\/div>\n<\/div>\n<\/div>\n
\n
8<\/div>
Question 8<\/div>
ICSE Class 10 Mathematics 2022 Semester 1<\/div><\/div><\/div>\n
\n
8<\/div>

Mohit opened a Recurring deposit account in a bank for 2 years. He deposits \u20b91000 every month and receives \u20b925500 on maturity. The interest he earned in 2 years is [1]<\/strong><\/p><\/div><\/div>\n

\n
A<\/span>\u20b913500<\/span><\/div>\n
B<\/span>\u20b93000<\/span><\/div>\n
C<\/span>\u20b924000<\/span><\/div>\n
D<\/span>\u20b91500<\/span><\/div>\n<\/div>\n
\u25bc<\/span> View Answer & Solution<\/div>\n
Answer & Solution<\/div>

Ans. Option (d) is correct.<\/p>\n

Explanation:<\/strong> <\/p>\n

\\[\n\\begin{aligned}\n\\text { Maturity Amount } & =\\text{\u20b9} 25500 \\\\\n\\text { Total deposit } & =1000 \\times 24=\\text{\u20b9} 24000 \\\\\n\\therefore \\quad \\text { Interest } & =\\text{\u20b9} 25,500-\\text{\u20b9} 24,000 \\\\\n& =\\text{\u20b9} 1,500\n\\end{aligned}\n\\]<\/div><\/div>\n<\/div>\n<\/div>\n
\n
9<\/div>
Question 9<\/div>
ICSE Class 10 Mathematics 2022 Semester 1<\/div><\/div><\/div>\n
\n
9<\/div>

In the given figure \\(\\boldsymbol{A} \\boldsymbol{B}=\\mathbf{2 4 ~ c m}, \\boldsymbol{A} \\boldsymbol{C}=\\mathbf{1 8 ~ c m}\\), \\(\\boldsymbol{D E}=\\mathbf{1 2 ~ c m}, \\boldsymbol{D F}=\\mathbf{9 ~ c m}\\) and \\(\\angle \\boldsymbol{B} A C=\\angle E D F\\). Then \\(\\triangle A B C \\sim \\triangle D E F\\) by the condition [1]<\/strong><\/p>\n

\"ICSE<\/div>\n
\"ICSE<\/div><\/div><\/div>\n
\n
A<\/span>\\(A A A\\)<\/span><\/div>\n
B<\/span>SAS<\/span><\/div>\n
C<\/span>\\(S S S\\)<\/span><\/div>\n
D<\/span>\\(A A S\\)<\/span><\/div>\n<\/div>\n
\u25bc<\/span> View Answer & Solution<\/div>\n
Answer & Solution<\/div>

Ans. Option (b) is correct.<\/p>\n

Explanation:<\/strong> In \\(\\triangle D E F\\) \\& \\(\\triangle A B C\\),<\/p>\n

\"ICSE<\/div>\n
\"ICSE<\/div>\n
\\[\n\\begin{aligned}\n\\frac{A C}{D F} & =\\frac{A B}{D E}=\\frac{2}{1} \\\\\n\\angle B A C & =\\angle E D F\n\\end{aligned}\n\\]<\/div>\n

Hence, \\(\\triangle A B C \\sim \\triangle D E F\\) (by SAS criterion)<\/p><\/div>\n<\/div>\n<\/div>\n

\n
10<\/div>
Question 10<\/div>
ICSE Class 10 Mathematics 2022 Semester 1<\/div><\/div><\/div>\n
\n
10<\/div>

If \\(\\boldsymbol{A}=\\left[\\begin{array}{cc}5 & 10 \\\\ 3 & -4\\end{array}\\right]\\) and \\(\\boldsymbol{I}=\\left[\\begin{array}{ll}1 & 0 \\\\ 0 & 1\\end{array}\\right]\\) then \\(\\boldsymbol{A} \\boldsymbol{I}\\) is equal to [1]<\/strong><\/p><\/div><\/div>\n

\n
A<\/span>\\(\\left[\\begin{array}{ll}1 & 0 \\\\ 0 & 1\\end{array}\\right]\\)<\/span><\/div>\n
B<\/span>\\(\\left[\\begin{array}{cc}5 & 10 \\\\ -3 & 4\\end{array}\\right]\\)<\/span><\/div>\n
C<\/span>\\(\\left[\\begin{array}{cc}5 & 10 \\\\ 3 & -4\\end{array}\\right]\\)<\/span><\/div>\n
D<\/span>\\(\\left[\\begin{array}{cc}15 & 15 \\\\ -1 & -1\\end{array}\\right]\\)<\/span><\/div>\n<\/div>\n
\u25bc<\/span> View Answer & Solution<\/div>\n
Answer & Solution<\/div>

Ans. Option (c) is correct.<\/p>\n

Explanation:<\/strong> \\(A \\times I=A\\)<\/p><\/div>\n<\/div>\n<\/div>\n

\n
11<\/div>
Question 11<\/div>
ICSE Class 10 Mathematics 2022 Semester 1<\/div><\/div><\/div>\n
\n
11<\/div>

The polynomial \\(\\boldsymbol{x}^{\\mathbf{3}}-\\mathbf{2} \\boldsymbol{x}^{\\mathbf{2}}+\\boldsymbol{a} \\boldsymbol{x}+\\mathbf{1 2}\\) when divided by \\((x+1)\\) leaves, remainder 20, then ‘ \\(a\\) ‘ is equal to [1]<\/strong><\/p><\/div><\/div>\n

\n
A<\/span>-31<\/span><\/div>\n
B<\/span>9<\/span><\/div>\n
C<\/span>11<\/span><\/div>\n
D<\/span>– 11<\/span><\/div>\n<\/div>\n
\u25bc<\/span> View Answer & Solution<\/div>\n
Answer & Solution<\/div>

Ans. Option (d) is correct.<\/p>\n

Explanation:<\/strong> <\/p>\n

\\[\n\\begin{aligned}\nx+1 & =0 \\\\\nx & =-1\n\\end{aligned}\n\\]<\/div>\n

Remainder \\(=p(-1)\\) Let<\/p>\n

\\[\np(x)=x^{3}-2 x^{2}+a x+12\n\\]<\/div>\n

Now, \\(p(-1)=20\\)<\/p>\n

\\[\n\\begin{array}{lr}\n\\Rightarrow & (-1)^{3}-2(-1)^{2}+a(-1)+12=20 \\\\\n\\Rightarrow & -1-2-a+12=20 \\\\\n\\Rightarrow & -a=20-9 \\\\\n\\Rightarrow & a=-11\n\\end{array}\n\\]<\/div><\/div>\n<\/div>\n<\/div>\n
\n
12<\/div>
Question 12<\/div>
ICSE Class 10 Mathematics 2022 Semester 1<\/div><\/div><\/div>\n
\n
12<\/div>

In an Arithmetic Progression (A.P.) if first term is 5, common difference is – 3 and the \\(\\boldsymbol{n}^{\\text {th }}\\) term is – 7, then \\(\\boldsymbol{n}\\) is equal to [1]<\/strong><\/p><\/div><\/div>\n

\n
A<\/span>5<\/span><\/div>\n
B<\/span>17<\/span><\/div>\n
C<\/span>– 13<\/span><\/div>\n
D<\/span>7<\/span><\/div>\n<\/div>\n
\u25bc<\/span> View Answer & Solution<\/div>\n
Answer & Solution<\/div>

Ans. Option (a) is correct.<\/p>\n

Explanation:<\/strong> Given, \\(a=5, d=-3\\) and \\(a_{n}=-7\\)<\/p>\n

\\[\n\\begin{array}{rlrl}\n\\text { Since, } & & a_{n} & =a+(n-1) d \\\\\n\\therefore & & -7 & =5+(n-1)(-3) \\\\\n\\Rightarrow & -7-5 & =-3 n+3 \\\\\n\\Rightarrow & -12-3 & =-3 n \\\\\n\\Rightarrow & -15 & =-3 n \\\\\n\\Rightarrow & n & =5\n\\end{array}\n\\]<\/div><\/div>\n<\/div>\n<\/div>\n
\n
13<\/div>
Question 13<\/div>
ICSE Class 10 Mathematics 2022 Semester 1<\/div><\/div><\/div>\n
\n
13<\/div>

In the given figure \\(\\boldsymbol{P} \\boldsymbol{Q}\\) is parallel to \\(\\boldsymbol{T} \\boldsymbol{R}\\), then by using condition of similarity [1]<\/strong><\/p>\n

\"ICSE<\/div><\/div><\/div>\n
\n
A<\/span>\\(\\frac{P Q}{R T}=\\frac{O P}{O T}=\\frac{O Q}{O R}\\)<\/span><\/div>\n
B<\/span>\\(\\frac{P Q}{R T}=\\frac{O P}{O R}=\\frac{O Q}{O T}\\)<\/span><\/div>\n
C<\/span>\\(\\frac{P Q}{R T}=\\frac{O R}{O P}=\\frac{O Q}{O T}\\)<\/span><\/div>\n
D<\/span>\\(\\frac{P Q}{R T}=\\frac{O P}{O R}=\\frac{O T}{O Q}\\)<\/span><\/div>\n<\/div>\n
\u25bc<\/span> View Answer & Solution<\/div>\n
Answer & Solution<\/div>

Ans. Option (b) is correct.<\/p>\n

Explanation:<\/strong> \\(P Q \\| T R, P R\\) is transversal,<\/p>\n

\"ICSE<\/div>\n
\\[\n\\Rightarrow \\quad \\angle P=\\angle R \\quad \\text { (alternate angles) }\n\\]<\/div>\n

Also, \\(Q T\\) is transversal,<\/p>\n

\\[\n\\begin{array}{lcc}\n\\Rightarrow & \\angle Q=\\angle T & \\text { (alternate angles) } \\\\\n\\therefore & \\triangle R T O \\sim \\triangle P Q O & \\text { (by AA criterion) }\n\\end{array}\n\\]<\/div>\n

\u2234<\/p>\n

\\[\n\\frac{P Q}{R T}=\\frac{P O}{R O}=\\frac{Q O}{T O}\n\\]<\/div>\n

(by C.P. C.T.)<\/p><\/div>\n<\/div>\n<\/div>\n

\n
14<\/div>
Question 14<\/div>
ICSE Class 10 Mathematics 2022 Semester 1<\/div><\/div><\/div>\n
\n
14<\/div>

If \\(\\boldsymbol{a}, \\boldsymbol{b}, \\boldsymbol{c}\\), and \\(\\boldsymbol{d}\\) are proportional then \\(\\frac{\\boldsymbol{a}+\\boldsymbol{b}}{\\boldsymbol{a}-\\boldsymbol{b}}\\) is equal to [1]<\/strong><\/p><\/div><\/div>\n

\n
A<\/span>\\(\\frac{c}{d}\\)<\/span><\/div>\n
B<\/span>\\(\\frac{c-d}{c+d}\\)<\/span><\/div>\n
C<\/span>\\(\\frac{d}{c}\\)<\/span><\/div>\n
D<\/span>\\(\\frac{c+d}{c-d}\\)<\/span><\/div>\n<\/div>\n
\u25bc<\/span> View Answer & Solution<\/div>\n
Answer & Solution<\/div>

Ans. Option (d) is correct.<\/p>\n

Explanation:<\/strong> Given,<\/p>\n

\\[\n\\frac{a}{b}=\\frac{c}{d}\n\\]<\/div>\n

By compodendo-dividendo<\/p>\n

\\[\n\\frac{a+b}{a-b}=\\frac{c+d}{c-d}\n\\]<\/div><\/div>\n<\/div>\n<\/div>\n
\n
15<\/div>
Question 15<\/div>
ICSE Class 10 Mathematics 2022 Semester 1<\/div><\/div><\/div>\n
\n
15<\/div>

The first four terms of an Arithmetic Progression (A. P.), whose first term is 4 and common difference is -6, are [1]<\/strong><\/p><\/div><\/div>\n

\n
A<\/span>4, – 10, – 16, – 22<\/span><\/div>\n
B<\/span>4, 10, 16, 22<\/span><\/div>\n
C<\/span>4, – 2, – 8, – 14<\/span><\/div>\n
D<\/span>4, 2, 8, 14<\/span><\/div>\n<\/div>\n
\u25bc<\/span> View Answer & Solution<\/div>\n
Answer & Solution<\/div>

Ans. Option (c) is correct.<\/p>\n

Explanation:<\/strong> \\(a=4 ; d=-6\\) First four terms of an A.P. are:<\/p>\n

\\[\n\\begin{array}{ll}\n& a, a+d, a+2 d, a+3 d \\\\\n\\Rightarrow & 4,4+(-6), 4+2(-6), 4+3(-6) \\\\\n\\Rightarrow & 4,-2,-8,-14\n\\end{array}\n\\]<\/div><\/div>\n<\/div>\n<\/div>\n
\n
16<\/div>
Question 16<\/div>
ICSE Class 10 Mathematics 2022 Semester 1<\/div><\/div><\/div>\n
\n
16<\/div>

One of the roots of the quadratic equation \\(\\boldsymbol{x}^{\\mathbf{2}}\\) – \\(8 x+5=0\\) is 7.3166. The root of the equation correct to 4 significant figures is [1]<\/strong><\/p><\/div><\/div>\n

\n
A<\/span>7.3166<\/span><\/div>\n
B<\/span>7.317<\/span><\/div>\n
C<\/span>7.316<\/span><\/div>\n
D<\/span>7.32<\/span><\/div>\n<\/div>\n
\u25bc<\/span> View Answer & Solution<\/div>\n
Answer & Solution<\/div>

Ans. Option (b) is correct.<\/p>\n

Explanation:<\/strong> Correct to 4 significant figures is 7.317.<\/p><\/div>\n<\/div>\n<\/div>\n

\n
17<\/div>
Question 17<\/div>
ICSE Class 10 Mathematics 2022 Semester 1<\/div><\/div><\/div>\n
\n
17<\/div>

\\((x+2)\\) and \\((x+3)\\) are two factors of the polynomial \\(\\boldsymbol{x}^{\\mathbf{3}}+\\mathbf{6} \\boldsymbol{x}^{\\mathbf{2}}+\\mathbf{1 1} \\boldsymbol{x}+\\mathbf{6}\\). If this polynomial is completely factorised the result is [2]<\/strong><\/p><\/div><\/div>\n

\n
A<\/span>\\((x-2)(x+3)(x+1)\\)<\/span><\/div>\n
B<\/span>\\((x+2)(x-3)(x-1)\\)<\/span><\/div>\n
C<\/span>\\((x+2)(x+3)(x-1)\\)<\/span><\/div>\n
D<\/span>\\((x+2)(x+3)(x+1)\\)<\/span><\/div>\n<\/div>\n
\u25bc<\/span> View Answer & Solution<\/div>\n
Answer & Solution<\/div>

Ans. Option (d) is correct.<\/p>\n

Explanation:<\/strong> Let<\/p>\n

\\[\np(x)=x^{3}+6 x^{2}+11 x+6\n\\]<\/div>\n

If<\/p>\n

\\[\nx+1=0\n\\]<\/div>\n

then<\/p>\n

\\[\nx=-1\n\\]<\/div>\n

Now,<\/p>\n

\\[\n\\begin{aligned}\np(-1) & =(-1)^{3}+6(-1)^{2}+11(-1)+6 \\\\\n& =-1+6-11+6=0\n\\end{aligned}\n\\]<\/div>\n

Hence, \\((x+1)\\) is a third factor.<\/p><\/div>\n<\/div>\n<\/div>\n

\n
18<\/div>
Question 18<\/div>
ICSE Class 10 Mathematics 2022 Semester 1<\/div><\/div><\/div>\n
\n
18<\/div>

The sum of the first 20 terms of the Arithmetic Progression 2, 4, 6, 8, ____ is [2]<\/strong><\/p><\/div><\/div>\n

\n
A<\/span>400<\/span><\/div>\n
B<\/span>840<\/span><\/div>\n
C<\/span>420<\/span><\/div>\n
D<\/span>800<\/span><\/div>\n<\/div>\n
\u25bc<\/span> View Answer & Solution<\/div>\n
Answer & Solution<\/div>

Ans. Option (c) is correct.<\/p>\n

Explanation:<\/strong> \\(S_{n}=\\frac{n}{2}[2 a+(n-1) d]\\) Here, \\(a=2, d=4-2=2\\) and \\(n=20\\) \u2234<\/p>\n

\\[\n\\begin{aligned}\nS_{20} & =\\frac{20}{2}[2 \\times 2+(20-1) 2] \\\\\n& =10[4+38] \\\\\n& =10 \\times 42=420\n\\end{aligned}\n\\]<\/div><\/div>\n<\/div>\n<\/div>\n
\n
19<\/div>
Question 19<\/div>
ICSE Class 10 Mathematics 2022 Semester 1<\/div><\/div><\/div>\n
\n
19<\/div>

The solution set on the number line of the linear inequation [2]<\/strong><\/p><\/div><\/div>\n

\n
A<\/span>
\"ICSE<\/div><\/span><\/div>\n
B<\/span>
\"ICSE<\/div><\/span><\/div>\n
C<\/span>
\"ICSE<\/div><\/span><\/div>\n
D<\/span>
\"ICSE<\/div><\/span><\/div>\n<\/div>\n
\u25bc<\/span> View Answer & Solution<\/div>\n
Answer & Solution<\/div>

Ans. Option (b) is correct.<\/p>\n

Explanation:<\/strong> <\/p>\n

\\[\n\\begin{aligned}\n2 y-6 & <y+2 \\\\\n2 y-y & <2+6 \\\\\ny & <8\n\\end{aligned}\n\\]<\/div>\n

and,<\/p>\n

\\[\n\\begin{aligned}\ny+2 & \\leq 2 y \\\\\n2 & \\leq 2 y-y \\\\\n2 & \\leq y\n\\end{aligned}\n\\]<\/div>\n

So,<\/p>\n

\\[\n2 \\leq y<8\n\\]<\/div><\/div>\n<\/div>\n<\/div>\n
\n
20<\/div>
Question 20<\/div>
ICSE Class 10 Mathematics 2022 Semester 1<\/div><\/div><\/div>\n
\n
20<\/div>

If \\(\\boldsymbol{x}, \\boldsymbol{y}, \\boldsymbol{z}\\) are in continued proportion then \\(\\left(\\boldsymbol{y}^{\\mathbf{2}}+\\boldsymbol{z}^{\\mathbf{2}}\\right)\\) : \\(\\left(x^{2}+y^{2}\\right)\\) is equal to [2]<\/strong><\/p><\/div><\/div>\n

\n
A<\/span>\\(z: x\\)<\/span><\/div>\n
B<\/span>\\(x: z\\)<\/span><\/div>\n
C<\/span>\\(z x\\)<\/span><\/div>\n
D<\/span>\\((y+z):(x+y)\\)<\/span><\/div>\n<\/div>\n
\u25bc<\/span> View Answer & Solution<\/div>\n
Answer & Solution<\/div>

Ans. Option (a) is correct.<\/p>\n

Explanation:<\/strong> \\(x, y, z\\) are in proportion<\/p>\n

\\[\ny^{2}=x z\n\\]<\/div>\n

\u2234<\/p>\n

\\[\n\\frac{y^{2}+z^{2}}{x^{2}+y^{2}}=\\frac{x z+z^{2}}{x^{2}+x z}=\\frac{z(x+z)}{x(x+z)}=\\frac{z}{x}\n\\]<\/div>\n

i.e.,<\/p>\n

\\[\nz: x\n\\]<\/div><\/div>\n<\/div>\n<\/div>\n
\n
21<\/div>
Question 21<\/div>
ICSE Class 10 Mathematics 2022 Semester 1<\/div><\/div><\/div>\n
\n
21<\/div>

The marked price of an article is \u20b9 5000. The shopkeeper gives a discount of 10%. If the rate of GST is 12%, then the amount paid by the customer including GST is [2]<\/strong><\/p><\/div><\/div>\n

\n
A<\/span>\u20b95040<\/span><\/div>\n
B<\/span>\u20b96100<\/span><\/div>\n
C<\/span>\u20b96272<\/span><\/div>\n
D<\/span>\u20b96160<\/span><\/div>\n<\/div>\n
\u25bc<\/span> View Answer & Solution<\/div>\n
Answer & Solution<\/div>

Ans. Option (a) is correct.<\/p>\n

Explanation:<\/strong> <\/p>\n

\\[\n\\begin{aligned}\n\\mathrm{SP} & =\\mathrm{MP}-\\text { Discount } \\\\\n& =5000-\\frac{10}{100} \\times 5000 \\\\\n& =5000-500=4500 \\\\\n\\text { GST } & =12 \\%\n\\end{aligned}\n\\]<\/div>\n
\\[\n\\text { GST amount }=\\frac{12}{100} \\times 4500=540\n\\]<\/div>\n

Total amount paid \\(=4500+540=\\text{\u20b9} 5040\\)<\/p><\/div>\n<\/div>\n<\/div>\n

\n
22<\/div>
Question 22<\/div>
ICSE Class 10 Mathematics 2022 Semester 1<\/div><\/div><\/div>\n
\n
22<\/div>

If \\(\\boldsymbol{A}=\\left[\\begin{array}{ll}3 & 5 \\\\ 1 & 4\\end{array}\\right], \\boldsymbol{B}=\\left[\\begin{array}{ll}2 & 4 \\\\ 0 & 3\\end{array}\\right]\\) and \\(\\boldsymbol{C}=\\left[\\begin{array}{cc}1 & -1 \\\\ 2 & 1\\end{array}\\right]\\), then \\(\\mathbf{5} \\boldsymbol{A}-\\boldsymbol{B} \\boldsymbol{C}\\) is equal to [2]<\/strong><\/p><\/div><\/div>\n

\n
A<\/span>\\(\\left[\\begin{array}{cc}-5 & -23 \\\\ 1 & 17\\end{array}\\right]\\)<\/span><\/div>\n
B<\/span>\\(\\left[\\begin{array}{cc}5 & 23 \\\\ 1 & 17\\end{array}\\right]\\)<\/span><\/div>\n
C<\/span>\\(\\left[\\begin{array}{ll}-2 & 8 \\\\ -3 & 3\\end{array}\\right]\\)<\/span><\/div>\n
D<\/span>\\(\\left[\\begin{array}{cc}5 & 23 \\\\ -1 & 17\\end{array}\\right]\\)<\/span><\/div>\n<\/div>\n
\u25bc<\/span> View Answer & Solution<\/div>\n
Answer & Solution<\/div>

Ans. Option (d) is correct.<\/p>\n

Explanation:<\/strong> <\/p>\n

\\[\n\\begin{aligned}\n5 A-B C & =5\\left[\\begin{array}{ll}\n3 & 5 \\\\\n1 & 4\n\\end{array}\\right]-\\left[\\begin{array}{ll}\n2 & 4 \\\\\n0 & 3\n\\end{array}\\right]\\left[\\begin{array}{cc}\n1 & -1 \\\\\n2 & 1\n\\end{array}\\right] \\\\\n& =\\left[\\begin{array}{cc}\n15 & 25 \\\\\n5 & 20\n\\end{array}\\right]-\\left[\\begin{array}{cc}\n10 & 2 \\\\\n6 & 3\n\\end{array}\\right] \\\\\n& =\\left[\\begin{array}{cc}\n5 & 23 \\\\\n-1 & 17\n\\end{array}\\right]\n\\end{aligned}\n\\]<\/div><\/div>\n<\/div>\n<\/div>\n
\n
23<\/div>
Question 23<\/div>
ICSE Class 10 Mathematics 2022 Semester 1<\/div><\/div><\/div>\n

In the given figure \\(A B C D\\) is a trapezium in which \\(D C\\) is parallel to \\(A B\\). \\(A B=\\mathbf{1 6 ~ c m}\\) and \\(D C=\\mathbf{8 ~ c m}, O D=\\mathbf{5 ~ c m}\\), \\(O B=(\\boldsymbol{y}+\\mathbf{3}) \\mathbf{~ c m}, \\boldsymbol{O} \\boldsymbol{A}=\\mathbf{1 1 ~ c m}\\) and \\(O C=(\\boldsymbol{x}-\\mathbf{1}) \\mathbf{c m}\\). Using the given information answer the following questions.<\/p>\n

\"ICSE<\/div><\/div>\n
\n
i<\/div>

From the given figure name the pair of similar triangles [1]<\/strong><\/p><\/div><\/div>\n

\n
A<\/span>\\(\\triangle O A B, \\triangle O B C\\)<\/span><\/div>\n
B<\/span>\\(\\triangle C O D, \\triangle A O B\\)<\/span><\/div>\n
C<\/span>\\(\\triangle A D B, \\triangle A C B\\)<\/span><\/div>\n
D<\/span>\\(\\triangle C O D, \\triangle C O B\\)<\/span><\/div>\n<\/div>\n
\u25bc<\/span> View Answer & Solution<\/div>\n
Answer & Solution<\/div>
\"ICSE<\/div>\n

Option (b) is correct.<\/p>\n

Explanation:<\/strong> \\(D C \\| A B\\) (given)<\/p>\n

\\[\n\\begin{array}{rll}\n& \\angle C D O & =\\angle A B O \\\\\n\\therefore D C O & =\\angle B A O & \\text { (alternate angle) } \\\\\n\\therefore & \\triangle C O D \\sim \\triangle A O B & \\text { (alternate angle) } A A \\text { criterion) }\n\\end{array}\n\\]<\/div><\/div>\n<\/div>\n
\n
ii<\/div>

The corresponding proportional sides with respect to the pair of similar triangles obtained in (i) [1]<\/strong><\/p><\/div><\/div>\n

\n
A<\/span>\\(\\frac{C D}{A B}=\\frac{O C}{O A}=\\frac{O D}{O B}\\)<\/span><\/div>\n
B<\/span>\\(\\frac{A D}{B C}=\\frac{O C}{O A}=\\frac{O D}{O B}\\)<\/span><\/div>\n
C<\/span>\\(\\frac{A D}{B C}=\\frac{B D}{A C}=\\frac{A B}{D C}\\)<\/span><\/div>\n
D<\/span>\\(\\frac{O D}{O B}=\\frac{C D}{C B}=\\frac{O C}{O A}\\)<\/span><\/div>\n<\/div>\n
\u25bc<\/span> View Answer & Solution<\/div>\n
Answer & Solution<\/div>

Option (a) is correct.<\/p>\n

Explanation:<\/strong> <\/p>\n

\\[\n\\triangle C O D \\sim \\triangle A O B\n\\]<\/div>\n

\u2234<\/p>\n

\\[\n\\frac{C D}{A B}=\\frac{O D}{O B}=\\frac{C O}{A O} \\text { (by C.P. C.T.) }\n\\]<\/div><\/div>\n<\/div>\n
\n
iii<\/div>

The ratio of the sides of the pair of similar triangles is [1]<\/strong><\/p><\/div><\/div>\n

\n
A<\/span>1 : 3<\/span><\/div>\n
B<\/span>1 : 2<\/span><\/div>\n
C<\/span>2 : 3<\/span><\/div>\n
D<\/span>3 : 1<\/span><\/div>\n<\/div>\n
\u25bc<\/span> View Answer & Solution<\/div>\n
Answer & Solution<\/div>

Option (b) is correct.<\/p>\n

Explanation:<\/strong> <\/p>\n

\\[\n\\frac{C D}{A B}=\\frac{O D}{O B}=\\frac{C O}{A O}=\\frac{8}{16}=\\frac{1}{2}\n\\]<\/div><\/div>\n<\/div>\n
\n
iv<\/div>

Using the ratio of sides of the pair of similar triangles the values of \\(\\boldsymbol{x}\\) and \\(\\boldsymbol{y}\\) are respectively [1]<\/strong><\/p><\/div><\/div>\n

\n
A<\/span>\\(x=4.6, y=7\\)<\/span><\/div>\n
B<\/span>\\(x=7, y=7\\)<\/span><\/div>\n
C<\/span>\\(x=6.5, y=7\\)<\/span><\/div>\n
D<\/span>\\(x=6.5, y=2\\)<\/span><\/div>\n<\/div>\n
\u25bc<\/span> View Answer & Solution<\/div>\n
Answer & Solution<\/div>

Option (c) is correct.<\/p>\n

Explanation:<\/strong> Here, \\(\\frac{x-1}{11}=\\frac{5}{y+3}=\\frac{1}{2}\\)<\/p>\n

\\[\n\\begin{aligned}\n& \\therefore \\frac{x-1}{11}=\\frac{1}{2} \\text { and } \\frac{5}{y+3}=\\frac{1}{2} \\\\\n& \\Rightarrow 2 x-2=11 \\text { and } y+3=10 \\\\\n& \\Rightarrow \\quad x=\\frac{13}{2}=6.5 \\text { and } y=7\n\\end{aligned}\n\\]<\/div><\/div>\n<\/div>\n<\/div>\n
\n
24<\/div>
Question 24<\/div>
ICSE Class 10 Mathematics 2022 Semester 1<\/div><\/div><\/div>\n

Two cars X and Y use 1 litre of diesel to travel \\(\\boldsymbol{x} \\mathbf{~ k m}\\) and \\((\\boldsymbol{x}+\\mathbf{3}) \\mathbf{k m}\\) respectively. If both the cars covered a distance of 72 km, then:<\/p><\/div>\n

\n
i<\/div>

The number of litres of diesel used by car X is [1]<\/strong><\/p><\/div><\/div>\n

\n
A<\/span>\\(\\frac{72}{x-3}\\) litres<\/span><\/div>\n
B<\/span>\\(\\frac{72}{x+3}\\) litres<\/span><\/div>\n
C<\/span>\\(\\frac{72}{x}\\) litres<\/span><\/div>\n
D<\/span>\\(\\frac{12}{x}\\) litres<\/span><\/div>\n<\/div>\n
\u25bc<\/span> View Answer & Solution<\/div>\n
Answer & Solution<\/div>

Option (c) is correct.<\/p>\n

Explanation:<\/strong> <\/p>\n

\\[\n\\begin{aligned}\nx \\mathrm{~km} & =1 \\text { litre } \\\\\n72 \\mathrm{~km} & =\\frac{72}{x} \\text { litres }\n\\end{aligned}\n\\]<\/div><\/div>\n<\/div>\n
\n
ii<\/div>

The number of litres of diesel used by car Y is [1]<\/strong><\/p><\/div><\/div>\n

\n
A<\/span>\\(\\frac{72}{x-3}\\) litres<\/span><\/div>\n
B<\/span>\\(\\frac{72}{x+3}\\) litres<\/span><\/div>\n
C<\/span>\\(\\frac{72}{x}\\) litres<\/span><\/div>\n
D<\/span>\\(\\frac{12}{x+3}\\) litres<\/span><\/div>\n<\/div>\n
\u25bc<\/span> View Answer & Solution<\/div>\n
Answer & Solution<\/div>

Option (b) is correct.<\/p>\n

Explanation:<\/strong> <\/p>\n

\\[\n\\begin{aligned}\n(x+3) \\mathrm{km} & =1 \\text { litre } \\\\\n72 \\mathrm{~km} & =\\frac{72}{x+3} \\text { litres }\n\\end{aligned}\n\\]<\/div><\/div>\n<\/div>\n
\n
iii<\/div>

If car X used 4 litres of diesel more than car Y in the journey, then [1]<\/strong><\/p><\/div><\/div>\n

\n
A<\/span>\\(\\frac{72}{x-3}-\\frac{12}{x}=4\\)<\/span><\/div>\n
B<\/span>\\(\\frac{72}{x+3}-\\frac{72}{x}=4\\)<\/span><\/div>\n
C<\/span>\\(\\frac{72}{x}-\\frac{72}{x+3}=4\\)<\/span><\/div>\n
D<\/span>\\(\\frac{72}{x-3}-\\frac{72}{x+3}=4\\)<\/span><\/div>\n<\/div>\n
\u25bc<\/span> View Answer & Solution<\/div>\n
Answer & Solution<\/div>

Option (c) is correct.<\/p>\n

Explanation:<\/strong> Since, given car X used 4 litres of diesel more than car Y. Therefore,<\/p>\n

\\[\n\\frac{72}{x}-\\frac{72}{x+3}=4\n\\]<\/div><\/div>\n<\/div>\n
\n
iv<\/div>

The amount of diesel used by the car X is [1]<\/strong><\/p><\/div><\/div>\n

\n
A<\/span>6 litres<\/span><\/div>\n
B<\/span>12 litres<\/span><\/div>\n
C<\/span>18 litres<\/span><\/div>\n
D<\/span>24 litres<\/span><\/div>\n<\/div>\n
\u25bc<\/span> View Answer & Solution<\/div>\n
Answer & Solution<\/div>

Option (b) is correct.<\/p>\n

Explanation:<\/strong> We have, \\(\\frac{72}{x}-\\frac{72}{x+3}=4\\)<\/p>\n

\\[\n\\begin{array}{lrl}\n\\Rightarrow & 72(x+3)-72 x & =4 x^{2}+12 x \\\\\n\\Rightarrow & 72 x+216-72 x & =4 x^{2}+12 x \\\\\n\\Rightarrow & 4 x^{2}+12 x-216 & =0 \\\\\n\\Rightarrow & x^{2}+3 x-54 & =0 \\\\\n\\Rightarrow & x^{2}+9 x-6 x-54 & =0 \\\\\n\\Rightarrow & x(x+9)-6(x+9) & =0 \\\\\n\\Rightarrow & (x-6)(x+9) & =0 \\\\\n\\Rightarrow & x & =6 \\\\\n& & \\text { (distance cannot be -ve) }\n\\end{array}\n\\]<\/div>\n

\u2234 Amount of diesel used by car X is \\(\\frac{72}{6}=12\\) litres<\/p><\/div>\n<\/div>\n<\/div>\n

\n
25<\/div>
Question 25<\/div>
ICSE Class 10 Mathematics 2022 Semester 1<\/div><\/div><\/div>\n

Joseph has a recurring deposit account in a bank for two years at the rate of 8%per annum simple interest.<\/p><\/div>\n

\n
i<\/div>

If at the time of maturity Joseph receives \u20b9 2000 as interest then the monthly installment is [1]<\/strong><\/p><\/div><\/div>\n

\n
A<\/span>\u20b9 1200<\/span><\/div>\n
B<\/span>\u20b9 600<\/span><\/div>\n
C<\/span>\u20b9 1000<\/span><\/div>\n
D<\/span>\u20b9 1600<\/span><\/div>\n<\/div>\n
\u25bc<\/span> View Answer & Solution<\/div>\n
Answer & Solution<\/div>

Option (c) is correct.<\/p>\n

Explanation:<\/strong> Given, \\(R=8 \\% ; I=\\text{\u20b9} 2000, n=24\\)<\/p>\n

\\[\n\\begin{aligned}\nI & =P \\times \\frac{n(n+1)}{2} \\times \\frac{R}{1200} \\\\\n2000 & =P \\times \\frac{24 \\times 25}{2} \\times \\frac{8}{1200} \\\\\n2000 & =P \\times 12 \\times \\frac{8}{48} \\\\\nP & =\\frac{2000}{2}=\\text{\u20b9} 1,000\n\\end{aligned}\n\\]<\/div><\/div>\n<\/div>\n
\n
ii<\/div>

The total amount deposited in the bank [1]<\/strong><\/p><\/div><\/div>\n

\n
A<\/span>\u20b9 25000<\/span><\/div>\n
B<\/span>\u20b9 24000<\/span><\/div>\n
C<\/span>\u20b9 26000<\/span><\/div>\n
D<\/span>\u20b9 23000<\/span><\/div>\n<\/div>\n
\u25bc<\/span> View Answer & Solution<\/div>\n
Answer & Solution<\/div>

Option (b) is correct.<\/p>\n

\\[\n\\begin{aligned}\n\\text { Explanation: Total amount } & =\\text{\u20b9} 1000 \\times 24 \\\\\n& =\\text{\u20b9} 24,000\n\\end{aligned}\n\\]<\/div><\/div>\n<\/div>\n
\n
iii<\/div>

The amount Joseph receives on maturity is [1]<\/strong><\/p><\/div><\/div>\n

\n
A<\/span>\u20b9 27000<\/span><\/div>\n
B<\/span>\u20b9 25000<\/span><\/div>\n
C<\/span>\u20b9 26000<\/span><\/div>\n
D<\/span>\u20b9 28000<\/span><\/div>\n<\/div>\n
\u25bc<\/span> View Answer & Solution<\/div>\n
Answer & Solution<\/div>

Option (c) is correct.<\/p>\n

Explanation:<\/strong> <\/p>\n

\\[\n\\begin{aligned}\n\\text { Maturity amount } & =\\text { Total amount }+ \\text { Interest } \\\\\n& =\\text{\u20b9} 24000+\\text{\u20b9} 2000 \\\\\n& =\\text{\u20b9} 26,000\n\\end{aligned}\n\\]<\/div><\/div>\n<\/div>\n
\n
iv<\/div>

If the monthly installment is \u20b9 100 and the rate of interest is 8%, in how many months Joseph will receive \u20b9 52 as interest? [1]<\/strong><\/p><\/div><\/div>\n

\n
A<\/span>18<\/span><\/div>\n
B<\/span>30<\/span><\/div>\n
C<\/span>12<\/span><\/div>\n
D<\/span>6<\/span><\/div>\n<\/div>\n
\u25bc<\/span> View Answer & Solution<\/div>\n
Answer & Solution<\/div>

Option (c) is correct.<\/p>\n

Explanation:<\/strong> \\(R=8 \\% ; I=\\text{\u20b9} 52, P=\\text{\u20b9} 100\\)<\/p>\n

\\[\n\\begin{aligned}\nI & =P \\times \\frac{n(n+1)}{2} \\times \\frac{R}{1200} \\\\\n52 & =100 \\times \\frac{n(n+1)}{2} \\times \\frac{8}{1200} \\\\\nn(n+1) & =3 \\times 52\n\\end{aligned}\n\\]<\/div>\n
\\[\n\\begin{array}{r}\nn^{2}+n-156=0 \\\\\nn^{2}+13 n-12 n-156=0 \\\\\nn(n+13)-12(n+13)=0\n\\end{array}\n\\]<\/div>\n
\\[\n\\begin{aligned}\n& \\therefore(n+13)(n-12)=0 \\\\\n& \\therefore n=12 \\text { months } \\\\\n& \\\\\n& \\square \\square \\square\n\\end{aligned}\n\\]<\/div><\/div>\n<\/div>\n<\/div>\n<\/div>\n\n","protected":false},"excerpt":{"rendered":"

Reading Time: <\/span> 22<\/span> minutes<\/span><\/span> ICSE Class 10 Mathematics PYQ Practice the complete ICSE Class 10 Mathematics 2022 Semester 1 Question Paper with Solutions. This […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_lmt_disableupdate":"","_lmt_disable":"","site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"default","ast-page-background-enabled":"default","ast-page-background-meta":{"desktop":{"background-color":"var(--ast-global-color-4)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"ast-content-background-meta":{"desktop":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"iawp_total_views":2,"footnotes":""},"categories":[6],"tags":[],"class_list":["post-2421","post","type-post","status-publish","format-standard","hentry","category-class-10-icse"],"modified_by":"sanjjeett","_links":{"self":[{"href":"https:\/\/xamcontent.com\/cisce\/wp-json\/wp\/v2\/posts\/2421"}],"collection":[{"href":"https:\/\/xamcontent.com\/cisce\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/xamcontent.com\/cisce\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/xamcontent.com\/cisce\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/xamcontent.com\/cisce\/wp-json\/wp\/v2\/comments?post=2421"}],"version-history":[{"count":1,"href":"https:\/\/xamcontent.com\/cisce\/wp-json\/wp\/v2\/posts\/2421\/revisions"}],"predecessor-version":[{"id":2423,"href":"https:\/\/xamcontent.com\/cisce\/wp-json\/wp\/v2\/posts\/2421\/revisions\/2423"}],"wp:attachment":[{"href":"https:\/\/xamcontent.com\/cisce\/wp-json\/wp\/v2\/media?parent=2421"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/xamcontent.com\/cisce\/wp-json\/wp\/v2\/categories?post=2421"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/xamcontent.com\/cisce\/wp-json\/wp\/v2\/tags?post=2421"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}