SSC GD Constable Maths Question and Answers download PDF based on previous year question paper of SSC GD exam. 40 Very important\u00a0 Maths questions for GD Constable.<\/p>\n
MATHS QUESTIONS FOR SSC GD PDF<\/a><\/p>\n GET 20 SSC GD MOCK FOR JUST RS. 117<\/a><\/p>\n FREE SSC EXAM YOUTUBE VIDEOS<\/a><\/p>\n Question 1:\u00a0<\/b>$7\\Large\\frac{1}{5}$ $+$ $2\\Large\\frac{1}{3}$ $-$ $4\\Large\\frac{2}{15}$ $-$ $3\\Large\\frac{5}{18}$ $+$ $5\\Large\\frac{2}{9}$ $=$?<\/p>\n a)\u00a0$5\\large\\frac{35}{78}$<\/p>\n b)\u00a0$7\\large\\frac{31}{90}$<\/p>\n c)\u00a0$9\\large\\frac{36}{73}$<\/p>\n d)\u00a0$4\\large\\frac{37}{79}$<\/p>\n e)\u00a0None of these<\/p>\n Question 2:\u00a0<\/b>$\\Large\\frac{628}{60}$ $-$ $\\Large\\frac{44}{45}\\div\\frac{6}{9}$ $+$ $\\Large\\frac{575}{624}\\div\\frac{25}{104}$ $+$ $\\Large\\frac{13}{15}\\div\\frac{2}{5}$ $=$ $\\large?$<\/p>\n a)\u00a015<\/p>\n b)\u00a025<\/p>\n c)\u00a010<\/p>\n d)\u00a014<\/p>\n e)\u00a0None of these<\/p>\n Question 3:\u00a0<\/b>$15.921$ $+$ $9.679$ $+$ $101.36$ $-$ $94.76$ $=$ $17.976$ $+$ $6.224$ $+$ ?<\/p>\n a)\u00a07<\/p>\n b)\u00a05<\/p>\n c)\u00a08<\/p>\n d)\u00a09<\/p>\n e)\u00a0None of these<\/p>\n Question 4:\u00a0<\/b>$\\sqrt{1936}\\div\\sqrt[3]{1331}\\times\\sqrt{50625}\\div\\sqrt{14400}\\times\\sqrt{64}$ $=$ $7920\\div132$ $+$ ?<\/p>\n a)\u00a01<\/p>\n b)\u00a0-3<\/p>\n c)\u00a00<\/p>\n d)\u00a0-1<\/p>\n e)\u00a0None of these<\/p>\n Question 5:\u00a0<\/b>$522\\div150\\times25$ $+$ $96\\div180\\times15$ $-$ $\\Large\\frac{1125}{135}$ $\\div$ $\\Large\\frac{5}{9}$ $=$ ?<\/p>\n a)\u00a080<\/p>\n b)\u00a075<\/p>\n c)\u00a0105<\/p>\n d)\u00a065<\/p>\n e)\u00a0None of these<\/p>\n SSC GD FREE PREVIOUS PAPERS<\/a><\/p>\n LATEST FREE SSC GD MOCK 2019<\/a><\/p>\n Question 6:\u00a0<\/b> \u092b\u0949\u0930\u094d\u092e \u0915\u0941\u0932\u094d\u0939\u093e\u0921\u093c\u0940 + + = = 0. \u0926\u094d\u0935\u093e\u0930\u093e \u090f\u0915 \u0938\u092e\u0940\u0915\u0930\u0923, \u091c\u0939\u093e\u0902, \u090f\u0915 \u2260 0, \u092c\u0940 \u2260 0 \u0914\u0930 \u0938\u0940 = 0 \u090f\u0915 \u0938\u0940\u0927\u0940 \u0930\u0947\u0916\u093e \u0915\u093e \u092a\u094d\u0930\u0924\u093f\u0928\u093f\u0927\u093f\u0924\u094d\u0935 \u0915\u0930\u0924\u093e \u0939\u0948 \u091c\u094b \u0917\u0941\u091c\u0930\u0924\u093e \u0939\u0948<\/p>\n a)\u00a0(2, 4)<\/p>\n b)\u00a0(0, 0)<\/p>\n c)\u00a0(3, 2)<\/p>\n d)\u00a0\u0907\u0928\u092e\u0947 \u0938\u0947 \u0915\u094b\u0908 \u0928\u0939\u0940\u0902<\/p>\n Question 7:\u00a0<\/b> \u0905\u0928\u0941\u0915\u094d\u0930\u092e \u0915\u093e \u092a\u093e\u0902\u091a\u0935\u093e\u0902 \u0936\u092c\u094d\u0926 \u091c\u093f\u0938\u0915\u0947 \u0932\u093f\u090f $t_{1}=1$ , $t_{2}=2$ \u0914\u0930 $t_{n+2}$ = $t_{n}+t_{n+1}$ , \u0939\u0948<\/p>\n a)\u00a0 5<\/p>\n b)\u00a0 10<\/p>\n c)\u00a0 6<\/p>\n d)\u00a0 8<\/p>\n Question 8:\u00a0<\/b> \u092f\u0926\u093f \u090f\u0915\u094d\u0938 = 0.3 $\\times$ 0.3, \u090f\u0915\u094d\u0938 \u0915\u093e \u092e\u093e\u0928 \u0939\u0948<\/p>\n a)\u00a00.009<\/p>\n b)\u00a00.03<\/p>\n c)\u00a00.09<\/p>\n d)\u00a00.08<\/p>\n Question 9:\u00a0<\/b>What would come in place of ($) mark in the following equation ? a)\u00a06<\/p>\n b)\u00a02<\/p>\n c)\u00a04<\/p>\n d)\u00a00<\/p>\n Question 10:\u00a0<\/b>If $\\ x^{2}+9y^{2}\\ $= 6xy, then x: y is<\/p>\n a)\u00a01 : 3<\/p>\n b)\u00a03 : 2<\/p>\n c)\u00a03 : 1<\/p>\n d)\u00a02 : 3<\/p>\n DOWNLOAD APP TO ACESS DIRECTLY ON MOBILE<\/a><\/p>\n Question 11:\u00a0<\/b>Which value among $3^{200},\\ 2^{300}\\ and\\ 7^{100}$ is the largest?<\/p>\n a)\u00a0$3^{200}$<\/p>\n b)\u00a0$2^{300}$<\/p>\n c)\u00a0$7^{100}$<\/p>\n d)\u00a0All are equal<\/p>\n Question 12:\u00a0<\/b>In a class, there are 40 students. Some of them passed the examination and others failed. Raman\u2019s rank among the student who have passed is 13 th from top and 17 th from bottom. How many students have failed?<\/p>\n a)\u00a011<\/p>\n b)\u00a010<\/p>\n c)\u00a09<\/p>\n d)\u00a0Cannot be determined<\/p>\n Question 13:\u00a0<\/b>By interchanging which two signs the equation will be correct? a)\u00a0+ and \u00f7<\/p>\n b)\u00a0+ and –<\/p>\n c)\u00a0\u00f7 and –<\/p>\n d)\u00a0None of these<\/p>\n Question 14:\u00a0<\/b>If 85 x 5 – 3 = 20 and 18 x 2 – 1 = 10, then 100 x 20 – 5 = ?<\/p>\n a)\u00a015<\/p>\n b)\u00a020<\/p>\n c)\u00a010<\/p>\n d)\u00a013<\/p>\n Question 15:\u00a0<\/b>If \u2018P\u2019 means \u2018+\u2019, \u2018Q\u2019 means \u2018-\u2019, \u2018R\u2019 means \u2018\u00f7\u2019 and \u2018S\u2019 means \u2018x\u2019, then which of the following equation is correct?<\/p>\n a)\u00a014 R 7 S 6 P 4 Q 3 = 11<\/p>\n b)\u00a03 S 6 P 2 Q 3 R 6 = 35\/2<\/p>\n c)\u00a011 R 12 S 48 P 10 Q 6 = 48<\/p>\n d)\u00a09 S 8 P 6 R 4 S 8 = 80<\/p>\n FREE SSC PRACTICE SET (DAILY TEST)<\/a><\/p>\n SSC CGL Free Mock Test<\/a><\/p>\n Question 16:\u00a0<\/b>By interchanging which two signs the equation will be correct? a)\u00a0+ and –<\/p>\n b)\u00a0– and \u00f7<\/p>\n c)\u00a0\u00f7 and x<\/p>\n d)\u00a0x and +<\/p>\n Question 17:\u00a0<\/b>If 13 L 4 A 7 = 41 and 14 A 3 L 12 = 54, then 12 L 3 A 9 = ?<\/p>\n a)\u00a084<\/p>\n b)\u00a039<\/p>\n c)\u00a042<\/p>\n d)\u00a056<\/p>\n Question 18:\u00a0<\/b>Which of the following statement is\/are true ? I. $\\sqrt{144}\\times\\sqrt{36}<\\sqrt[3]{125}\\times\\sqrt{121}$ a)\u00a0Only I<\/p>\n b)\u00a0Only II<\/p>\n c)\u00a0Neither I nor II<\/p>\n d)\u00a0Both I and II<\/p>\n Question 19:\u00a0<\/b>Which of the following is the correct rationalize form of $\\frac{15}{\\sqrt{5}+2}=?$<\/p>\n a)\u00a0$5\\sqrt{5}-6$<\/p>\n b)\u00a0$5\\sqrt{5}-30$<\/p>\n c)\u00a0$15\\sqrt{5}-30$<\/p>\n d)\u00a0$45\\sqrt{5}-630$<\/p>\n Question 20:\u00a0<\/b>If $\\sqrt[3]{7}+\\sqrt{343}=19.21$, then find the value of $\\sqrt{252}+20\\sqrt{7}$<\/p>\n a)\u00a039.4<\/p>\n b)\u00a049.9<\/p>\n c)\u00a056.8<\/p>\n d)\u00a092.3<\/p>\n Question 21:\u00a0<\/b>Find the least number among $\\frac{5}{9},\\sqrt\\frac{9}{49},0.43$ and $(0.7)^{2}.$<\/p>\n a)\u00a0$\\frac{9}{5}$<\/p>\n b)\u00a0$(0.7)^{2}$<\/p>\n c)\u00a0$\\sqrt\\frac{9}{49}$<\/p>\n d)\u00a00.43<\/p>\n Question 22:\u00a0<\/b>What is the value of $(\\frac{\\sqrt{2}}{3}-Cosec 60^\\circ)?$<\/p>\n a)\u00a0$\\frac{(\\sqrt{6}-6)}{3\\sqrt{3}}$<\/p>\n b)\u00a0$\\frac{(2-2\\sqrt{3})}{\\sqrt{3}}$<\/p>\n c)\u00a0$\\frac{(1-\\sqrt{6})}{\\sqrt{2}}$<\/p>\n d)\u00a0$\\frac{(4-\\sqrt{3})}{2\\sqrt{3}}$<\/p>\n Question 23:\u00a0<\/b>What is the value of $\\sqrt{1054 +\\sqrt{196}+\\sqrt{169}+\\sqrt{64}}$ ?<\/p>\n a)\u00a033<\/p>\n b)\u00a037<\/p>\n c)\u00a029<\/p>\n d)\u00a031<\/p>\n Question 24:\u00a0<\/b>What is the positive square root of $[19+4\\sqrt21]?$<\/p>\n a)\u00a0$\\sqrt{7}+2\\sqrt{3}$<\/p>\n b)\u00a0$\\sqrt{3}+2\\sqrt{7}$<\/p>\n c)\u00a0$\\sqrt{2}+3\\sqrt{7}$<\/p>\n d)\u00a0$\\sqrt{7}+3\\sqrt{3}$<\/p>\n Question 25:\u00a0<\/b>Find ‘x’ if $\\sqrt{(2+7x)}=\\sqrt{(3x+4)}$<\/p>\n a)\u00a00.5<\/p>\n b)\u00a01<\/p>\n c)\u00a01.5<\/p>\n d)\u00a02<\/p>\n Question 26:\u00a0<\/b>What is the square root of $\\frac{(3-2\\sqrt2)}{(3+2\\sqrt2)}$ ?<\/p>\n a)\u00a0$3-2\\sqrt2$<\/p>\n b)\u00a0$3+2\\sqrt2$<\/p>\n c)\u00a0$1$<\/p>\n d)\u00a0$17$<\/p>\n Question 27:\u00a0<\/b>What is the square root of $\\frac{(3-2\\sqrt2)}{(3+2\\sqrt2)}?$<\/p>\n a)\u00a0$3-2\\sqrt2$<\/p>\n b)\u00a0$3+2\\sqrt2$<\/p>\n c)\u00a0$1$<\/p>\n d)\u00a0$$17$<\/p>\n Question 28:\u00a0<\/b>Find the value of $\\sqrt{(2x-5)^{2}}+2\\sqrt{(x-1)^{2}}$, if 1<x<2.<\/p>\n a)\u00a01<\/p>\n b)\u00a02<\/p>\n c)\u00a03<\/p>\n d)\u00a04<\/p>\n Question 29:\u00a0<\/b>What is the value of $(\\frac{1}{\\sqrt{3}}+\\cos60^\\circ)$ ?<\/p>\n a)\u00a0$\\frac{(2+2\\sqrt3)}{\\sqrt{3}}$<\/p>\n b)\u00a0$\\frac{(2+\\sqrt3)}{2\\sqrt{3}}$<\/p>\n c)\u00a0$\\frac{7}{3}$<\/p>\n d)\u00a0$\\frac{(\\sqrt{2}+1)}{\\sqrt{2}}$<\/p>\n Question 30:\u00a0<\/b>If $x=\\frac{\\sqrt2+1}{\\sqrt2-1}$ and $y=\\frac{\\sqrt2-1}{\\sqrt2+1}$, then what is the value of x+y ?<\/p>\n a)\u00a0$6$<\/p>\n b)\u00a0$\\sqrt2$<\/p>\n c)\u00a0$3$<\/p>\n d)\u00a0$3\\sqrt2$<\/p>\n Question 31:\u00a0<\/b>What is the simplified value of $\\frac{\\sqrt{5}+\\sqrt{4}}{\\sqrt{5}-\\sqrt{4}}+\\frac{\\sqrt{5}-\\sqrt{4}}{\\sqrt{5}+\\sqrt{4}}$ ?<\/p>\n a)\u00a09<\/p>\n b)\u00a010<\/p>\n c)\u00a018<\/p>\n d)\u00a020<\/p>\n Question 32:\u00a0<\/b>What is value of $(6x^2 – 5y^2)(6x^2 + 5y^2)$, If $x=\\frac{1}{\\sqrt{3}}$ and $y=\\frac{1}{\\sqrt{5}}$ ?<\/p>\n a)\u00a02<\/p>\n b)\u00a03<\/p>\n c)\u00a04<\/p>\n d)\u00a05<\/p>\n Question 33:\u00a0<\/b>What is the value of $(cosec 60^\\circ – \\frac{1}{2})$<\/p>\n a)\u00a0$\\frac{(4-\\sqrt{3})}{2\\sqrt{3}}$<\/p>\n b)\u00a0$\\frac{(2\\sqrt{3}-1)}{\\sqrt{3}}$<\/p>\n c)\u00a0$\\frac{(3\\sqrt{3}-1)}{3}$<\/p>\n d)\u00a0$\\frac{1}{\\sqrt{3}}$<\/p>\n Question 34:\u00a0<\/b>If $\\sqrt{21}=4.58$, then what is the simplified value of $(8\\sqrt{\\frac{3}{7}}-3\\sqrt{\\frac{7}{3}})$ ?<\/p>\n a)\u00a00.474<\/p>\n b)\u00a00.752<\/p>\n c)\u00a00.655<\/p>\n d)\u00a01<\/p>\n Question 35:\u00a0<\/b>Arrange the following in descending order. a)\u00a0$\\sqrt[4]{6}>\\sqrt[3]{4}>\\sqrt[4]{5}$<\/p>\n b)\u00a0$\\sqrt[4]{6}>\\sqrt[4]{5}>\\sqrt[3]{4}$<\/p>\n c)\u00a0$\\sqrt[3]{4}>\\sqrt[4]{5}>\\sqrt[4]{6}$<\/p>\n d)\u00a0$\\sqrt[3]{4}>\\sqrt[4]{6}>\\sqrt[4]{5}$<\/p>\n Question 36:\u00a0<\/b>Which of the following relation is CORRECT? a)\u00a0Only I<\/p>\n b)\u00a0Only II<\/p>\n c)\u00a0Neither I nor II<\/p>\n d)\u00a0Both I and II<\/p>\n Question 37:\u00a0<\/b>Which of the following statement (s) is \/ are TRUE<\/strong> ? a)\u00a0Only I<\/p>\n b)\u00a0Only II<\/p>\n c)\u00a0Neither I nor II<\/p>\n d)\u00a0Both I and II<\/p>\n Question 38:\u00a0<\/b>Determine the value of m for which $4x+\\frac{\\sqrt{x}}{6}+\\frac{m^2}{4}$ is a perfect square.<\/p>\n a)\u00a0$\\frac{1}{24}$<\/p>\n b)\u00a0$\\frac{1}{12}$<\/p>\n c)\u00a0$12$<\/p>\n d)\u00a0$24$<\/p>\n Question 39:\u00a0<\/b>Calculate the value of x, If $\\sqrt{1-(\\frac{x}{529}})=(\\frac{16}{23})$<\/p>\n a)\u00a0283<\/p>\n b)\u00a0276<\/p>\n c)\u00a0273<\/p>\n d)\u00a0374<\/p>\n Question 40:\u00a0<\/b>If X=1.1, then what is the value of $\\sqrt{4x^2-4x+1}$ ?<\/p>\n a)\u00a01.21<\/p>\n b)\u00a01.331<\/p>\n c)\u00a02.21<\/p>\n d)\u00a01.2<\/p>\n Answers & Solutions:<\/strong><\/span><\/p>\n 1)\u00a0Answer\u00a0(B)<\/strong><\/p>\n 7$\\frac{1}{5}$+2$\\frac{1}{3}$-4$\\frac{2}{15}$-3$\\frac{5}{18}$+5$\\frac{2}{9}$<\/p>\n = 7+2-4-3+5+$\\frac{1}{5}$+$\\frac{1}{3}$-$\\frac{2}{15}$-$\\frac{5}{18}$+$\\frac{2}{9}$<\/p>\n =7+ $\\frac{18+30-12-25+20}{90}$<\/p>\n = 7$\\frac{31}{90}$<\/p>\n 2)\u00a0Answer\u00a0(A)<\/strong><\/p>\n $\\Large\\frac{628}{60}$ $-$ $\\Large\\frac{44}{45}\\div\\frac{6}{9}$ $+$ $\\Large\\frac{575}{624}\\div\\frac{25}{104}$ $+$ $\\Large\\frac{13}{15}\\div\\frac{2}{5}$<\/p>\n $=$ $\\Large\\frac{314}{30}$ $-$ $\\Large\\frac{44}{30}$ $+$ $\\Large\\frac{115}{30}$ $+$ $\\Large\\frac{65}{30}$<\/p>\n $=$ $\\Large\\frac{450}{30}$ $=15$<\/p>\n 3)\u00a0Answer\u00a0(C)<\/strong><\/p>\n $15.921$ $+$ $9.679$ $+$ $101.36$ $-$ $94.76$ $-$ $17.976$ $-$ $6.224$ $=$ ?<\/p>\n Adding all Positive terms: Adding all Negative terms: $126.96$ $-$ $118.96$ $=$ $8$<\/p>\n 4)\u00a0Answer\u00a0(C)<\/strong><\/p>\n $\\large\\sqrt{1936}\\div\\sqrt[3]{1331}\\times\\sqrt{50625}\\div\\sqrt{14400}\\times\\sqrt{64}$ $-$ $\\large7920\\div132$ $=$ ?<\/p>\n $\\Large\\frac{44}{11}$ $\\times$ $\\Large\\frac{225}{(\\Large\\frac{120}{8})}$ $-$ $\\Large\\frac{7920}{132}$<\/p>\n $4$ $\\times$ $15$ $-$ $60$ $=$ $0$<\/p>\n 5)\u00a0Answer\u00a0(A)<\/strong><\/p>\n $522\\div150\\times25$ $=$ $87$<\/p>\n $96\\div180\\times15$ $=$ $8$<\/p>\n $\\Large\\frac{1125}{135}$ $\\div$ $\\Large\\frac{5}{9}$ $=$ $15$<\/p>\n $87$ $+$ $8$ $-$ $15$ $=$ $80$<\/p>\n SSC GD FREE STUDY MATERIAL<\/a><\/p>\n 6)\u00a0Answer\u00a0(B)<\/strong><\/p>\n \u0938\u0940 = 0 \u0915\u0947 \u0930\u0942\u092a \u092e\u0947\u0902, \u0914\u0930 \u0938\u092e\u0940\u0915\u0930\u0923 \u092e\u0947\u0902 \u092c\u093f\u0902\u0926\u0941 (0,0) \u0915\u094b \u092a\u094d\u0930\u0924\u093f\u0938\u094d\u0925\u093e\u092a\u093f\u0924 \u0915\u0930\u0928\u0947 \u0915\u0947 \u0932\u093f\u090f, \u0939\u092e\u0947\u0902 \u092c\u093f\u0902\u0926\u0941 (0,0) \u092a\u0930 ax + by + c = 0 \u092e\u093f\u0932\u0924\u093e \u0939\u0948\u0964 7)\u00a0Answer\u00a0(D)<\/strong><\/p>\n $t_{1}=1$ , $t_{2}=2$<\/p>\n $t_{n+2}$ = $t_{n}+t_{n+1}$<\/p>\n n = 3 $t_{5}$ , \u092b\u093f\u0930 $t_{5}$ = $t_{3}+t_{4}$<\/p>\n $t_{3}$ = $t_{1}+t_{2}$ = 1 + 2 = 3<\/p>\n $t_{4}$ = $t_{2}+t_{3}$ = 2 + 3 = 5<\/p>\n $t_{5}$ = $t_{3}+t_{4}$ = 3 + 5 = 8<\/p>\n \u0924\u094b \u091c\u0935\u093e\u092c \u0935\u093f\u0915\u0932\u094d\u092a \u0921\u0940 \u0939\u0948\u0964<\/p>\n 8)\u00a0Answer\u00a0(C)<\/strong><\/p>\n 0.3 * 0.3 = 0.0 9<\/p>\n 9)\u00a0Answer\u00a0(A)<\/strong><\/p>\n 10)\u00a0Answer\u00a0(C)<\/strong><\/p>\n $x^{2}+9y^{2}$=6xy<\/p>\n $\\Rightarrow x^{2}-6xy+9y^{2}$=0<\/p>\n $\\Rightarrow (x-3y)^{2}$=0<\/p>\n $\\Rightarrow$ x-3y=0<\/p>\n $\\Rightarrow$ x=3y<\/p>\n $\\Rightarrow \\frac{x}{y}$=$\\frac{3}{1}$<\/p>\n $\\therefore$ x:y=3:1<\/p>\n 11)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Terms = $3^{200},\\ 2^{300}\\ and\\ 7^{100}$<\/p>\n Dividing all the exponents by 100, we get :<\/p>\n $\\equiv3^2,2^3,7^1$<\/p>\n = $9,8,7$<\/p>\n Thus, the largest number = $9\\equiv3^{200}$<\/p>\n => Ans – (A)<\/p>\n 12)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Total number of students = 40<\/p>\n Among the student who have passed, Raman’s rank from top = 13th<\/p>\n Raman’s rank from bottom = 17th<\/p>\n => Total students who passed = $(13+17)-1=30-1=29$<\/p>\n $\\therefore$ Number of students who have failed = $40-29=11$<\/p>\n => Ans – (A)<\/p>\n 13)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Expression : 25 + 18 \u00f7 2 – 4 = 20<\/p>\n (A) : + and \u00f7<\/p>\n $\\equiv25\\div18+2-4=20$<\/p>\n L.H.S. = $1.39-2=-0.61\\neq20$<\/p>\n (B) : + and –<\/p>\n $\\equiv25-18\\div2+4=20$<\/p>\n L.H.S. = $25-9+4=20$<\/p>\n => Ans – (B)<\/p>\n 14)\u00a0Answer\u00a0(C)<\/strong><\/p>\n The pattern followed is $\\times$ is replaced by $\\div$ and $-$ is replaced by $+$<\/p>\n Eg :- 85 x 5 – 3 = $\\frac{85}{5}+3=17+3=20$<\/p>\n and 18 x 2 – 1 = $\\frac{18}{2}+1=9+1=10$<\/p>\n Similarly, 100 x 20 – 5 = $\\frac{100}{20}+5=5+5=10$<\/p>\n => Ans – (C)<\/p>\n 15)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Given : \u2018P\u2019 means \u2018+\u2019, \u2018Q\u2019 means \u2018-\u2019, \u2018R\u2019 means \u2018\u00f7\u2019 and \u2018S\u2019 means \u2018x\u2019<\/p>\n (A) : 14 R 7 S 6 P 4 Q 3 = 11<\/p>\n L.H.S. = $14\\div7\\times6+4-3$<\/p>\n = $12+1=13\\neq$ R.H.S.<\/p>\n (B) : 3 S 6 P 2 Q 3 R 6 = 35\/2<\/p>\n L.H.S. = $3\\times6+2-3\\div6$<\/p>\n = $18+2-0.5=19.5=\\frac{39}{2}\\neq$ R.H.S.<\/p>\n (C) : 11 R 12 S 48 P 10 Q 6 = 48<\/p>\n L.H.S. = $11\\div12\\times48+10-6$<\/p>\n = $44+4=48=$ R.H.S.<\/p>\n => Ans – (C)<\/p>\n 16)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Expression : 16 + 31 – 3 x 93 \u00f7 11 = 966<\/p>\n By exchanging :<\/p>\n (A) + and –<\/p>\n L.H.S. = $16-31+3\\times93\\div11$<\/p>\n = $-15+25.36=10.36\\neq$ R.H.S.<\/p>\n (B) – and \u00f7<\/p>\n L.H.S. = $16+31\\div3\\times93-11$<\/p>\n = $16+(31\\times31)-11$<\/p>\n = $5+961=966=$ R.H.S.<\/p>\n => Ans – (B)<\/p>\n 17)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Expression : 13 L 4 A 7 = 41<\/p>\n The pattern followed is : $A\\rightarrow\\times$ and $L\\rightarrow+$<\/p>\n => $13+(4\\times7)=41$<\/p>\n and Similarly, $14\\times3+12=54$<\/p>\n $\\therefore$ 12 L 3 A 9<\/p>\n = $12+3\\times9=12+27=39$<\/p>\n => Ans – (B)<\/p>\n 18)\u00a0Answer\u00a0(B)<\/strong><\/p>\n I : $\\sqrt{144}\\times\\sqrt{36}<\\sqrt[3]{125}\\times\\sqrt{121}$<\/p>\n L.H.S. = $12\\times6=72$<\/p>\n R.H.S. = $5\\times11=55$<\/p>\n Thus, L.H.S. > R.H.S.<\/p>\n II : $\\sqrt{324}+\\sqrt{49}>\\sqrt[3]{216}\\times\\sqrt{9}$<\/p>\n L.H.S. = $18+7=25$<\/p>\n R.H.S. = $6\\times3=18$<\/p>\n Thus, L.H.S. > R.H.S., which is correct.<\/p>\n $\\therefore$ Only II is correct.<\/p>\n => Ans – (B)<\/p>\n 19)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Expression : $\\frac{15}{\\sqrt{5}+2}=?$<\/p>\n Rationalizing the denominator, we get :<\/p>\n = $\\frac{15}{\\sqrt{5}+2}\\times\\frac{(\\sqrt5-2)}{(\\sqrt5-2)}$<\/p>\n = $\\frac{15(\\sqrt5-2)}{5-4}=15\\sqrt5-30$<\/p>\n => Ans – (C)<\/p>\n 20)\u00a0Answer\u00a0(B)<\/strong><\/p>\n 21)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Terms : $\\frac{5}{9}\\approx0.56$<\/p>\n $\\sqrt{\\frac{9}{49}}$ $=\\frac{3}{7}\\approx0.42$<\/p>\n $0.43$<\/p>\n $(0.7)^2=0.49$<\/p>\n Thus, the least number is = $\\sqrt\\frac{9}{49}$<\/p>\n => Ans – (C)<\/p>\n 22)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Expression : $(\\frac{\\sqrt{2}}{3}-Cosec 60^\\circ)$<\/p>\n = $\\frac{\\sqrt2}{3}-\\frac{2}{\\sqrt3}$<\/p>\n = $\\frac{(\\sqrt{6}-6)}{3\\sqrt{3}}$<\/p>\n => Ans – (A)<\/p>\n 23)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Expression : $\\sqrt{1054+ \\sqrt{196}+\\sqrt{169}+\\sqrt{64}}$<\/p>\n = $\\sqrt{1054+14+13+8}$<\/p>\n = $\\sqrt{1089}=33$<\/p>\n => Ans – (A)<\/p>\n 24)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Expression : $[19+4\\sqrt21]?$<\/p>\n = $[19+2\\sqrt{4\\times21}]=[19+2\\sqrt{84}]$<\/p>\n = $7+12+2\\sqrt{7\\times12}$<\/p>\n = $(7)^2+(12)^2+2\\sqrt{7\\times12}$<\/p>\n Now, we know that $a^2+b^2+2ab=(a+b)^2$<\/p>\n = $(\\sqrt{7}+\\sqrt{12})^2$<\/p>\n Thus, square root is = $\\sqrt{7}+\\sqrt{12}$<\/p>\n = $\\sqrt7+2\\sqrt3$<\/p>\n => Ans – (A)<\/p>\n 25)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Expression : $\\sqrt{(2+7x)}=\\sqrt{(3x+4)}$<\/p>\n Squaring both sides, we get :<\/p>\n => $2+7x=3x+4$<\/p>\n => $7x-3x=4-2$<\/p>\n => $4x=2$<\/p>\n => $x=\\frac{2}{4}=0.5$<\/p>\n => Ans – (A)<\/p>\n 26)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Given : $x=\\frac{(3-2\\sqrt2)}{(3+2\\sqrt2)}$<\/p>\n To find : $\\sqrt{x}$<\/p>\n Solution : rationalizing the denominator, we get<\/p>\n => $\\frac{(3-2\\sqrt2)}{(3+2\\sqrt2)}\\times\\frac{(3-2\\sqrt2)}{(3-2\\sqrt2)}$<\/p>\n = $\\frac{(3-2\\sqrt2)^2}{(3)^2-(2\\sqrt2)^2}$<\/p>\n = $\\frac{(3-2\\sqrt2)^2}{9-8}$<\/p>\n => $x=(3-2\\sqrt2)^2$<\/p>\n Taking square root on both sides,<\/p>\n => $\\sqrt{x}=3-2\\sqrt2$<\/p>\n => Ans – (A)<\/p>\n 27)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Expression : $\\frac{(3-2\\sqrt2)}{(3+2\\sqrt2)}$<\/p>\n Rationalizing the denominator,<\/p>\n = $\\frac{(3-2\\sqrt2)}{(3+2\\sqrt2)}\\times\\frac{(3-2\\sqrt2)}{(3-2\\sqrt2)}$<\/p>\n = $\\frac{(3-2\\sqrt2)^2}{(3+2\\sqrt2)(3-2\\sqrt2)}$<\/p>\n = $\\frac{(3-2\\sqrt2)^2}{9-8}=(3-2\\sqrt2)^2$<\/p>\n Thus, square root is = $3-2\\sqrt2$<\/p>\n => Ans – (A)<\/p>\n 28)\u00a0Answer\u00a0(C)<\/strong><\/p>\n 29)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Expression : $(\\frac{1}{\\sqrt{3}}+\\cos60^\\circ)$<\/p>\n = $\\frac{1}{\\sqrt3}+\\frac{1}{2}$<\/p>\n = $\\frac{(2+\\sqrt3)}{2\\sqrt{3}}$<\/p>\n => Ans – (B)<\/p>\n 30)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Given : $x=\\frac{\\sqrt2+1}{\\sqrt2-1}$ and $y=\\frac{\\sqrt2-1}{\\sqrt2+1}$<\/p>\n To find : $x+y=$ $\\frac{\\sqrt2+1}{\\sqrt2-1}+\\frac{\\sqrt2-1}{\\sqrt2+1}$<\/p>\n = $\\frac{(\\sqrt2+1)^2+(\\sqrt2-1)^2}{(\\sqrt2-1)(\\sqrt2+1)}$<\/p>\n = $\\frac{(2+1+2\\sqrt{2})+(2+1-2\\sqrt{2})}{2-1}$<\/p>\n = $\\frac{(3+3)}{1}=6$<\/p>\n => Ans – (A)<\/p>\n 31)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Expression : $\\frac{\\sqrt{5}+\\sqrt{4}}{\\sqrt{5}-\\sqrt{4}}+\\frac{\\sqrt{5}-\\sqrt{4}}{\\sqrt{5}+\\sqrt{4}}$<\/p>\n = $\\frac{(\\sqrt5+\\sqrt4)^2+(\\sqrt5-\\sqrt4)^2}{(\\sqrt5-\\sqrt4)(\\sqrt5+\\sqrt4)}$<\/p>\n = $\\frac{(5+4+2\\sqrt{20})+(5+4-2\\sqrt{20})}{5-4}$<\/p>\n = $\\frac{(9+9)}{1}=18$<\/p>\n => Ans – (C)<\/p>\n 32)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Given : $x=\\frac{1}{\\sqrt{3}}$ and $y=\\frac{1}{\\sqrt{5}}$<\/p>\n => $x^2=\\frac{1}{3}$ and $y^2=\\frac{1}{5}$<\/p>\n Again, squaring both terms, we get : $x^4=\\frac{1}{9}$ and $y^4=\\frac{1}{25}$ ————-(i)<\/p>\n To find : $(6x^2 – 5y^2)(6x^2 + 5y^2)$<\/p>\n Using, $(a-b)(a+b)=a^2-b^2$ and substituting values from equation (i),<\/p>\n = $36x^4-25y^4$<\/p>\n = $(36\\times\\frac{1}{9})-(25\\times\\frac{1}{25})$<\/p>\n = $4-1=3$<\/p>\n => Ans – (B)<\/p>\n 33)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Expression : $(cosec 60^\\circ – \\frac{1}{2})$<\/p>\n = $\\frac{2}{\\sqrt3}-\\frac{1}{2}$<\/p>\n = $\\frac{(4-\\sqrt{3})}{2\\sqrt{3}}$<\/p>\n => Ans – (A)<\/p>\n 34)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Given : $\\sqrt{21}=4.58$<\/p>\n To find : $(8\\sqrt{\\frac{3}{7}}-3\\sqrt{\\frac{7}{3}})$<\/p>\n = $\\frac{(8\\times3)-(3\\times7)}{\\sqrt{21}}$<\/p>\n = $\\frac{(24-21)}{\\sqrt{21}}=\\frac{3}{\\sqrt{21}}$<\/p>\n = $\\frac{3}{4.58}\\approx0.655$<\/p>\n => Ans – (C)<\/p>\n 35)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Terms : $\\sqrt[4]{5},\\sqrt[3]{4}$ and $\\sqrt[4]{6}$<\/p>\n Multiplying the exponents by L.C.M. (4,3,4) = 12<\/p>\n => $(5)^{\\frac{12}{4}}$ , $(4)^{\\frac{12}{3}}$ and $(6)^{\\frac{12}{4}}$<\/p>\n = $5^3,4^4,6^3$<\/p>\n = $125,256,216$<\/p>\n Thus, in descending order = $256>216>125$<\/p>\n $\\equiv$ $\\sqrt[3]{4}>\\sqrt[4]{6}>\\sqrt[4]{5}$<\/p>\n => Ans – (D)<\/p>\n 36)\u00a0Answer\u00a0(A)<\/strong><\/p>\n I : $(\\sqrt{15}+\\sqrt{7})<(2\\sqrt{22})$<\/p>\n Squaring both sides, we get :<\/p>\n L.H.S. = $(\\sqrt{15}+\\sqrt{7})^2=15+7+2\\sqrt{105}=(22+2\\sqrt{105})\\approx(22+2\\times10)=42$<\/p>\n R.H.S. = $(2\\sqrt{22})^2=88$<\/p>\n Thus, L.H.S. < R.H.S., which is correct.<\/p>\n II : $(\\sqrt{17}+\\sqrt{5})<(\\sqrt{20}+\\sqrt{2})$<\/p>\n Squaring both sides, we get :<\/p>\n L.H.S. = $(\\sqrt{17}+\\sqrt{5})^2=17+5+2\\sqrt{85}=(22+2\\sqrt{85})$<\/p>\n R.H.S. = $(\\sqrt{20}+\\sqrt{2})^2=20+2+2\\sqrt{40}=(22+2\\sqrt{40})$<\/p>\n $\\because$ $\\sqrt{85}>\\sqrt{40}$, then L.H.S. > R.H.S.<\/p>\n Thus, only I is correct.<\/p>\n => Ans – (A)<\/p>\n 37)\u00a0Answer\u00a0(B)<\/strong><\/p>\n I : $\\sqrt{676}+\\sqrt{6.76}+\\sqrt{0.0676}=27.76$<\/p>\n L.H.S. = $26+2.6+0.26=28.86\\neq$ R.H.S.<\/p>\n II : $\\sqrt{339+\\sqrt{36}+\\sqrt{49}+\\sqrt{81}}=19$<\/p>\n L.H.S. = $\\sqrt{339+6+7+9}=\\sqrt{361}=19=$ R.H.S.<\/p>\n Thus, only II is correct.<\/p>\n => Ans – (B)<\/p>\n 38)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Expression : $4x+\\frac{\\sqrt{x}}{6}+\\frac{m^2}{4}$<\/p>\n Let $\\sqrt{x}=y$<\/p>\n = $4y^2+\\frac{y}{6}+\\frac{m^2}{4}$<\/p>\n = $(2y)^2+2(2y)(\\frac{1}{24})+(\\frac{m}{2})^2$<\/p>\n Using, $a^2+2ab+b^2=(a+b)^2$<\/p>\n => $\\frac{m}{2}=\\frac{1}{24}$<\/p>\n => $m=\\frac{2}{24}=\\frac{1}{12}$<\/p>\n => Ans – (B)<\/p>\n 39)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Expression : $\\sqrt{1-(\\frac{x}{529}})=(\\frac{16}{23})$<\/p>\n => $\\sqrt{\\frac{529-x}{529}}=\\frac{16}{23}$<\/p>\n => $\\frac{529-x}{529}=(\\frac{16}{23})^2$<\/p>\n => $\\frac{529-x}{529}=\\frac{256}{529}$<\/p>\n => $529-x=256$<\/p>\n => $x=529-256=273$<\/p>\n => Ans – (C)<\/p>\n 40)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Expression : $\\sqrt{4x^2-4x+1}$<\/p>\n = $\\sqrt{(2x-1)^2}=(2x-1)$ ————-(i)<\/p>\n It is given that $x=1.1$, substituting it in equation (i),<\/p>\n => $2(1.1)-1=2.2-1=1.2$<\/p>\n => Ans – (D)<\/p>\n
\n* 2 $ 20 \u00f7 156 = 145<\/p>\n
\n25 + 18 \u00f7 2 – 4 = 20<\/p>\n
\n16 + 31 – 3 x 93 \u00f7 11 = 966<\/p>\n
\nII.$\\sqrt{324}+\\sqrt{49}>\\sqrt[3]{216}\\times\\sqrt{9}$<\/p>\n
\n$\\sqrt[4]{5},\\sqrt[3]{4}$ and $\\sqrt[4]{6}$<\/p>\n
\nI. $(\\sqrt{15}+\\sqrt{7})<(2\\sqrt{22})$
\nII. $(\\sqrt{17}+\\sqrt{5})<(\\sqrt{20}+\\sqrt{2})$<\/p>\n
\nI. $\\sqrt{676}+\\sqrt{6.76}+\\sqrt{0.0676}=27.76$
\nII. $\\sqrt{339+\\sqrt{36}+\\sqrt{49}+\\sqrt{81}}=19$<\/p>\n
\n$15.921$ $+$ $9.679$ $+$ $101.36$ $=$ $126.96$<\/p>\n
\n$94.76$ $+$ $17.976$ $+$ $6.224$ $=$ $118.96$<\/p>\n
\n\u0907\u0938\u0932\u093f\u090f, \u0930\u0947\u0916\u093e \u092e\u0942\u0932 \u0915\u0947 \u092e\u093e\u0927\u094d\u092f\u092e \u0938\u0947 \u0917\u0941\u091c\u0930\u0924\u0940 \u0939\u0948\u0964<\/p>\n