Download SBI Clerk Expected Quant Questions & Answers PDF for SBI Clerk Prelims and Mains exam. Very Important SBI Clerk Expected Quant questions with solutions.<\/p>\n
Download SBI Clerk Expected Quant Questions Set-2<\/a><\/p>\n 790+ Mocks – Just Rs. 194. Use coupon: SBIDREAM70<\/a><\/p>\n Download SBI Clerk Puzzles Set-1 PDF<\/a><\/p>\n Question 1:\u00a0<\/b>In 8 years, Ramesh will be twice as old as Farheen was 8 years ago. If Ramesh is now 16 years older than Farheen, the present age of Farheen is :<\/p>\n a)\u00a032<\/p>\n b)\u00a040<\/p>\n c)\u00a036<\/p>\n d)\u00a044<\/p>\n e)\u00a024<\/p>\n Question 2:\u00a0<\/b>A was thrice as old as B when C was 2 years old. A will be twice as old as B when C is 10 years old. How old will B be when the difference between the ages of A and B will be equal to the age of C?<\/p>\n a)\u00a016 years<\/p>\n b)\u00a022 years<\/p>\n c)\u00a018 years<\/p>\n d)\u00a024 years<\/p>\n e)\u00a032 years<\/p>\n Question 3:\u00a0<\/b>The ratio of P\u2019s age 3 years ago and Q\u2019s age after 5 years is 6 : 13. If at present, the ratio of their ages is 11 : 20, then what is the ratio of P\u2019s age 2 years hence and Q\u2019s age 5 years ago?<\/p>\n a)\u00a0$\\frac{9}{11}$<\/p>\n b)\u00a0$\\frac{7}{9}$<\/p>\n c)\u00a0$\\frac{5}{13}$<\/p>\n d)\u00a0$\\frac{7}{11}$<\/p>\n e)\u00a0$\\frac{8}{13}$<\/p>\n Download SBI Clerk Previous Papers PDF<\/a><\/p>\n Take a free mock test for SBI Clerk<\/a><\/p>\n Question 4:\u00a0<\/b>The sum of the ages of a P and Q is 53 years. Four years ago, the product of their ages was two times the square of Q\u2019s age. The present age of P is?<\/p>\n a)\u00a034 years<\/p>\n b)\u00a038 years<\/p>\n c)\u00a032 years<\/p>\n d)\u00a042 years<\/p>\n e)\u00a036 years<\/p>\n Question 5:\u00a0<\/b>The present age of Sanjay is 40 percent of his mother\u2019s present age. 8 years from now, he will be half as old as his mother. What is the sum of present ages of Sanjay and his mother?<\/p>\n a)\u00a042 years<\/p>\n b)\u00a070 years<\/p>\n c)\u00a056 years<\/p>\n d)\u00a084 years<\/p>\n e)\u00a050 years<\/p>\n Question 6:\u00a0<\/b>Find the approximate value of $\\Large\\frac{169.87^{2}-13.94^{2}}{183.87}$ $\\large+$ $\\Large\\frac{3211.96}{44.002}$<\/p>\n a)\u00a0224<\/p>\n b)\u00a0236<\/p>\n c)\u00a0239<\/p>\n d)\u00a0229<\/p>\n e)\u00a0None of these<\/p>\n Question 7:\u00a0<\/b>Find the approximate value of $17.33+142.895-76.795+235.008-4.779+38.102$<\/p>\n a)\u00a0321<\/p>\n b)\u00a0351<\/p>\n c)\u00a0281<\/p>\n d)\u00a0371<\/p>\n e)\u00a0None of these<\/p>\n Banking Study Material – 18000 Questions<\/a><\/p>\n Daily Free SBI Online Tests<\/a><\/p>\n Question 8:\u00a0<\/b>Find the approximate value of ‘x’ in $\\Large\\frac{383.96}{x} = \\frac{x}{24.01}$<\/p>\n a)\u00a084<\/p>\n b)\u00a096<\/p>\n c)\u00a092<\/p>\n d)\u00a088<\/p>\n e)\u00a0None of these<\/p>\n Question 9:\u00a0<\/b>Find the approximate value of $\\sqrt{6560}\\div2.98^{2}+\\sqrt[3]{2189}\\div\\sqrt{2700}\\times\\sqrt{5180}\\div\\sqrt[3]{25}$<\/p>\n a)\u00a011<\/p>\n b)\u00a015<\/p>\n c)\u00a020<\/p>\n d)\u00a05<\/p>\n e)\u00a0None of these<\/p>\n Question 10:\u00a0<\/b>Find the approximate value of $153.001\\times146.95-66.009\\times73.952+25.001\\times23.856$<\/p>\n a)\u00a017952<\/p>\n b)\u00a018207<\/p>\n c)\u00a018456<\/p>\n d)\u00a017605<\/p>\n e)\u00a0None of these<\/p>\n 200+ Free GK Tests<\/a><\/p>\n Question 11:\u00a0<\/b>The average marks of a class increased by $\\large\\frac{1}{3}$ when a student\u2019s marks were wrongly entered as 64 instead of 46. The number of students in the class are<\/p>\n a)\u00a054<\/p>\n b)\u00a058<\/p>\n c)\u00a063<\/p>\n d)\u00a057<\/p>\n e)\u00a0None of these<\/p>\n Question 12:\u00a0<\/b>The average age of students in a class is 12. If the average age of boys is 18 and the average age of girls is 14 and If the number of boys are 22, then find the total number of students in the class<\/p>\n a)\u00a052<\/p>\n b)\u00a055<\/p>\n c)\u00a060<\/p>\n d)\u00a058<\/p>\n e)\u00a0None of these<\/p>\n Question 13:\u00a0<\/b>In the first 20 overs of a cricket game, the run rate was 4.5, What should be the run rate in the next 30 overs of the game to chase the target of 210?<\/p>\n a)\u00a06<\/p>\n b)\u00a04<\/p>\n c)\u00a08<\/p>\n d)\u00a06.5<\/p>\n e)\u00a0None of these<\/p>\n Question 14:\u00a0<\/b>In a society, 60% of residents own a car and 40% own a bike. Nobody owns car and bike both. Also, 25% of the residents work in the banking sector. If 20% of those who work in the banking sector own a bike, then the residents who own a car and work in non-banking sector form what percentage of the total people?<\/p>\n a)\u00a025%<\/p>\n b)\u00a040%<\/p>\n c)\u00a037.5%<\/p>\n d)\u00a060%<\/p>\n e)\u00a042.5%<\/p>\n Question 15:\u00a0<\/b>Amit scored 10% less marks than Sumit and 20% more marks than Rakesh. If Tina scored 35 marks more than Amit and Sumit scored 20 marks less than Tina, what is the absolute difference between the marks of Tina and Rakesh?<\/p>\n a)\u00a075<\/p>\n b)\u00a082.5<\/p>\n c)\u00a037.5<\/p>\n d)\u00a057.5<\/p>\n e)\u00a040<\/p>\n Instructions<\/b><\/p>\n Select the correct option.<\/p>\n Question 16:\u00a0<\/b>Quantity 1:A number is increased by 85% and then it is decreased by 15% and result is approximately 357 a)\u00a0Quantity 1 $\\leq$Quantity 2<\/p>\n b)\u00a0Quantity 1 $\\geq$Quantity 2<\/p>\n c)\u00a0Quantity 1 < Quantity 2<\/p>\n d)\u00a0Quantity 1 > Quantity 2<\/p>\n e)\u00a0Quantity 1 = Quantity 2<\/p>\n Question 17:\u00a0<\/b>Quantity 1:If the cost price of total apples bought is Rs 288 and each apple is sold at Rs 14 each and the profit percent is 50\/3 then number of apples sold is a)\u00a0Quantity 1 $\\leq$Quantity 2<\/p>\n b)\u00a0Quantity 1 $\\geq$Quantity 2<\/p>\n c)\u00a0Quantity 1 < Quantity 2<\/p>\n d)\u00a0Quantity 1 > Quantity 2<\/p>\n e)\u00a0Quantity 1 = Quantity 2<\/p>\n Question 18:\u00a0<\/b>Quantity 1:Numerator of a fraction is increased by 20% an the denominator is decreased by 25% and the resultant is 2 then the sum of numerator and denominator of the original fraction. a)\u00a0Quantity 1 $\\leq$Quantity 2<\/p>\n b)\u00a0Quantity 1 $\\geq$Quantity 2<\/p>\n c)\u00a0Quantity 1 < Quantity 2<\/p>\n d)\u00a0Quantity 1 > Quantity 2<\/p>\n e)\u00a0Quantity 1 = Quantity 2<\/p>\n Question 19:\u00a0<\/b>Quantity 1:Roots of the equation $6x^{2}-13x+6$=0 a)\u00a0Quantity 1 $\\leq$Quantity 2<\/p>\n b)\u00a0Quantity 1 $\\geq$Quantity 2<\/p>\n c)\u00a0Quantity 1 < Quantity 2<\/p>\n d)\u00a0Quantity 1 > Quantity 2<\/p>\n e)\u00a0Quantity 1 = Quantity 2<\/p>\n Instructions<\/b><\/p>\n Calculate the quantity I and the quantity II on the basis of the given information then compare them and answer the following questions accordingly.<\/p>\n Question 20:\u00a0<\/b>Quantity I: In a party, each person shakes hand with every other person. If there are total 45 handshakes, then the total number of persons present are? a)\u00a0Quantity I > Quantity II<\/p>\n b)\u00a0Quantity I < Quantity II<\/p>\n c)\u00a0Quantity I $\\geq$ Quantity II<\/p>\n d)\u00a0Quantity I $\\leq$ Quantity II<\/p>\n e)\u00a0Quantity I = Quantity II or The relationship between Quantity I and Quantity II cannot be established.<\/p>\n General Knowledge Questions & Answers PDF<\/a><\/p>\n Answers & Solutions:<\/strong><\/span><\/p>\n 1)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Let us assume the present age of Farheen = x years Ramesh\u2019s age after 8 years will be = x+16+8 = x+24 years $\\Rightarrow$ x+24 = 2*(x-8) So the present age of Farheen will be 40 years.<\/p>\n 2)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Let the age of A and B initially be \u2018a\u2019 and \u2018b\u2019 respectively.<\/p>\n When C = 2, a\/b = 3 When C is as old as the difference between the ages of A and B, i.e, when C is 24-8 = 16 years old, age of B will be 8+14 = 22 years. Therefore, option B is the right answer.<\/p>\n 3)\u00a0Answer\u00a0(D)<\/strong><\/p>\n We have, Hence, option D is the right answer.<\/p>\n 4)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Let the age of P and Q be p and q respectively. 5)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Let the present age of Sanjay\u2019s mother be \u2018x\u2019. So Sanjay\u2019s present age will be .4x 6)\u00a0Answer\u00a0(D)<\/strong><\/p>\n $\\Large\\frac{169.87^{2}-13.94^{2}}{183.87}$ $\\large+$ $\\Large\\frac{3211.96}{44.002}\\simeq \\frac{170^{2}-14^{2}}{184}$ $\\large+$ $\\Large\\frac{3212}{44}$<\/p>\n $= \\Large\\frac{(170+14)(170-14)}{184}$ $+$ $73$<\/p>\n $= \\Large\\frac{184\\times156}{184}$ $+$ $73$<\/p>\n $= 156+73 = 229$<\/p>\n 7)\u00a0Answer\u00a0(B)<\/strong><\/p>\n $17.33+142.895-76.795+235.008-4.779+38.102 \\simeq 17+143-77+235-5+38$<\/p>\n Adding all positive terms: Adding all negative terms: $\\therefore$ $17+143-77+235-5+38 = 433-82 = 351$<\/p>\n 8)\u00a0Answer\u00a0(B)<\/strong><\/p>\n <\/p>\n $\\Large\\frac{383.96}{x} = \\frac{x}{24.01}$<\/p>\n $\\Rightarrow x^{2} \\simeq 384\\times24 = 9216$<\/p>\n $\\therefore$ x $= \\sqrt{9216} = 96$<\/p>\n 9)\u00a0Answer\u00a0(B)<\/strong><\/p>\n $\\sqrt{6560}\\div2.98^{2}+\\sqrt[3]{2189}\\div\\sqrt{2700}\\times\\sqrt{5180}\\div\\sqrt[3]{25} \\simeq \\sqrt{6561}\\div3^{2}+\\sqrt[3]{2197}\\div\\sqrt{2704}\\times\\sqrt{5184}\\div\\sqrt[3]{27}$<\/p>\n $= \\Large\\frac{\\sqrt{6561}}{3^{2}}+\\frac{\\sqrt[3]{2197}\\times\\sqrt{5184}}{\\sqrt{2704}\\times\\sqrt[3]{27}}$<\/p>\n $= \\Large\\frac{81}{9}+\\frac{13\\times72}{52\\times3}$ $= 9+6 = 15$<\/p>\n 10)\u00a0Answer\u00a0(B)<\/strong><\/p>\n $153.001\\times146.95-66.009\\times73.952+25.001\\times23.856 \\simeq 153\\times147-66\\times74+25\\times24$<\/p>\n $153\\times147$ can be written as $(150+3)(150-3)$<\/p>\n $=$ $150^{2}-3^{2}$ $(\\because (a+b)(a-b) = a^{2}-b^{2})$ $66\\times74$ can be written as $(70-4)(70+4)$ $=$ $70^{2}-4^{2}$ $=$ $4900-16 = 4884$<\/p>\n $25\\times24$ can be written as $24\\times\\Large\\frac{100}{4} =$ $600$<\/p>\n $\\therefore$ $153\\times147-66\\times74+25\\times24 = 22491-4884+600 = 18207$<\/p>\n 11)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Let the number of students be ‘x’<\/p>\n Given that the average is increased by $\\Large\\frac{1}{3}$ due to the increase of $18$ marks ( $64$ $-$ $46$ $=$ $18$ marks )<\/p>\n Actual increase in marks $=$ $\\Large\\frac{1}{3}$ $\\times$ x<\/p>\n $\\Rightarrow$ $\\Large\\frac{1}{3}$ $\\times$ x $=$ $18$<\/p>\n $\\therefore$ Number of students(x) $=$ $54$<\/p>\n 12)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Let the total age of boys be \u2018B\u2019<\/p>\n Average age of boys = 18<\/p>\n $\\Large\\frac{B}{22}$ $=$ $18$<\/p>\n $\\Rightarrow$ B $=$ $396$<\/p>\n Let the total age of Girls be ‘G’<\/p>\n Let the number of girls be ‘x’<\/p>\n Average age of girls $=$ $14$<\/p>\n $\\Large\\frac{G}{x}$ $=$ $14$<\/p>\n $\\Rightarrow$ G $=$ $14$x<\/p>\n Total average of students $=$ $12$<\/p>\n Total age of students $=$ Total age of boys $+$ Total age of girls $=$ $396$ $+$ $14$x<\/p>\n Total number of students $=$ Number of boys $+$ Number of girls $=$ $22$ $+$ x<\/p>\n Average age of class $=$ $12$<\/p>\n $\\Rightarrow$ $\\Large\\frac{396 + 14x}{22 + x}$ $=$ $12$<\/p>\n $\\Rightarrow$ $396$ $+$ $14$x $=$ $462$ $+$ $12$x<\/p>\n $\\Rightarrow$ $2$x $=$ $66$<\/p>\n $\\Rightarrow$ x $=$ $33$<\/p>\n $\\therefore$ Number of girls $=$ $33$<\/p>\n Total number of students $=$ $22$ $+$ $33$ $=$ $55$<\/p>\n 13)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Run-rate = 4.5 runs per over<\/p>\n $1$ over $\\rightarrow$ $4.5$ runs Total target $=$ $210$ runs<\/p>\n Remaining score $=$ $210$ $-$ $90$ $=$ $120$ runs<\/p>\n Remaining overs $=$ $30$ overs<\/p>\n $\\therefore$ Required run-rate $=$ $\\Large\\frac{120}{30}$ $=$ $4$ runs per over<\/p>\n 14)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Let the total number of residents be 100x 15)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Let Sumit scored 100x marks 16)\u00a0Answer\u00a0(D)<\/strong><\/p>\n In Q1 we have x*1.85*0.85=357 17)\u00a0Answer\u00a0(C)<\/strong><\/p>\n In Q1 we have Total CP=288 18)\u00a0Answer\u00a0(E)<\/strong><\/p>\n In Q1 let the fraction be be (x\/y) 19)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Roots of the equation in Q1 is $6x^{2}-9x-4x+6$=0 20)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Quantity I: \u21d2 n = 10 or n = -9 Quantity II: Therefore, Quantity I < Quantity II<\/p>\n
\nQuantity 2:A number is increased by 82% and then it is decreased by 12% and the result is approximately 355.5<\/p>\n
\nQuantity 2:An article is marked up by 40% an a discount of 30% was given then 13 times the loss percent is<\/p>\n
\nQuantity 2:Numerator of a fraction is increased by 50% and denominator is decreased by 60% and the resultant is 3 then the sum of numerator and denominator of original fraction is<\/p>\n
\nQuantity 2:Roots of the equation $25x^{2}-25x+6$=0<\/p>\n
\nQuantity II: In a party, each person shakes hand with every other person. If there are 8 persons, then the number of handshakes possible are?<\/p>\n
\nSo present age of Ramesh will be = x + 16 years<\/p>\n
\nFarheen\u2019s age 8 years ago = x – 8 years<\/p>\n
\n$\\Rightarrow$ x= 40 years<\/p>\n
\n=> a = 3b
\nWhen C = 10, 8 years have passed.
\n=> (a+8)\/(b+8) = 2
\na+8 = 2b + 16
\na = 2b+8
\n=> b = 8 years and a = 24 years.<\/p>\n
\n$\\frac{P – 3}{Q + 5} = \\frac{6}{13}$
\nSo, 13P – 39 = 6Q + 30
\n13P – 6Q = 69
\n$\\frac{P}{Q} = \\frac{11}{20}$
\n20P – 11Q = 0
\nSolving both the equations we have,
\nP = 33 years and Q = 60 years
\nSo, $\\frac{P + 2}{Q – 5} = \\frac{35}{55} = \\frac{7}{11}$<\/p>\n
\nWe have, p+q = 53
\n$(p-4)(q-4) = 2 * (q-4)^2$
\np-4 = 2q – 8
\np = 2q – 4
\nSubstituing the value of p in first equation,
\n3q – 4 = 53
\nq = 19
\np = 53 – 19 = 34
\nHence, option A is right choice.<\/p>\n
\nWe have been given that
\n(.4x + 8)*2 = x + 8
\n=> .8x + 16 = x + 8
\n=> .2x = 8
\n=> x = 40
\nHence, the present age of the mother is 40 years. Thus, required sum = 56.<\/p>\n
\n$17+143+235+38 = 433$<\/p>\n
\n$77+5 = 82$<\/p>\n
\n$=$ $22500-9$ $=$ $22491$<\/p>\n
\n$20$ overs $\\rightarrow$ $20\\times4.5$ $=$ $90$ runs<\/p>\n
\nPeople who own car = 60k
\nPeople who own a bike = 40k
\nPeople who work in the banking sector = 25k
\nPeople in the banking sector who own a bike = 20% of 25k = 5k
\nRest of the people who work in the banking sector must be owning a car
\nNumber of people owning a car and working in banking sector = 25k – 5k = 20k
\nNumber of people owning a car and working in non-banking sector = 60k – 20k = 40k
\nRequired % = 40k\/100k * 100 = 40%
\nHence, option B is the correct answer.<\/p>\n
\nThen, Amit scored 90x marks
\nAmit\u2019s score is 20% more than that of Rakesh.
\nSo, Rakesh\u2019s score = 75x
\nTina\u2019s score = Amit\u2019s score + 35 = 90x + 35…..(i)
\nTina\u2019s score = Sumit\u2019s score + 20 = 100x + 20…(ii)
\nOn comparing (i) and (ii),
\n90x + 35 = 100x + 20
\nOn solving, we get x = 1.5
\nSo, Rakesh\u2019s score = 75x = 75* 1.5 = 112.5
\nAnd Tina\u2019s score = 90x + 35 = 170
\nDifference = (170 – 112.5) = 57.5
\nHence, option D is the correct answer.<\/p>\n
\nx=357\/(1.85*0.85)
\nx=227
\nIn Q2 we have x*1.82*0.88=355.5
\nx=355.5\/(1.82*0.88)
\nx=220
\nTherefore Q1>Q2<\/p>\n
\nLet the number of apples be x
\nTotal SP=14x
\nTherefore ((14x-288)\/(288))*100=50\/3
\n14x-288=48
\n14x=288+48
\nx=336\/14
\nx=24
\nIn Q2 we have CP=100
\nMP=100*1.4=140
\nSP=MP-discount
\nSP=140-140*(30\/100)
\nSP=98
\nLoss percent=((100-98)\/(100))*100
\n=2%
\n13 times the loss percent=26
\nTherefore Q2 > Q1<\/p>\n
\nThen as given (1.2x)\/(0.75y)=2
\nx\/y =5\/4
\nsum=5+4=9
\nIn Q2 let the fraction be (x\/y)
\nThen as given (1.5x)\/(0.4y)=3
\nx\/y =4\/5
\nsum=4+5=9<\/p>\n
\n3x(2x-3)-2(2x-3)=0
\n(3x-2)(2x-3)=0
\nx=2\/3 and x=3\/2
\nIn Q2 we have $25x^{2}-15x-10x+6$=0
\n5x(5x-3)-2(5x-3)=0
\n(5x-2)(5x-3)=0
\nx=2\/5 and x=3\/5
\nAnd so by comparing Q1 > Q2<\/p>\n
\nNumber of handshakes possible = $\\dfrac{n(n-1)}{2}$ where n = number of persons
\nGiven, $\\dfrac{n(n-1)}{2} = 45$
\n\u21d2 $n^2 – n = 90$
\n\u21d2 $n^1 – n – 90 = 0$
\n\u21d2 $n^2 – 10n + 9n – 90 = 0$
\n\u21d2 $n(n-10) + 9(n-10) = 0$
\n\u21d2 $(n-10)(n+9) = 0$<\/p>\n
\nNumber of persons cannot be negative.
\nHence, Total number of persons is 10.<\/p>\n
\nNumber of handshakes possible = $\\dfrac{n(n-1)}{2}$ where n = number of persons.
\n= $\\dfrac{8 \\times 7}{2} = 28$<\/p>\n