35 IBPS RRB PO Mocks @ Rs. 149. Enroll Now<\/a><\/p>\n 70 IBPS RRB (PO + Clerk) Mocks @ Rs. 199<\/a><\/p>\n Take a free mock test for IBPS RRB PO<\/a><\/p>\n Download IBPS RRB PO Previous Papers PDF<\/a><\/p>\n Question 1:\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 2:\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 Question 3:\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 4:\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 5:\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 Question 6:\u00a0<\/b>\u00a0Find the approximate value of $\\small1798.98\\div8.01\\times51.965\\div779.97-390.001\\div25.87$<\/p>\n a)\u00a05<\/p>\n b)\u00a012<\/p>\n c)\u00a00<\/p>\n d)\u00a020<\/p>\n e)\u00a0None of these<\/p>\n Free Mock Test for IBPS RRB PO<\/a><\/p>\n IBPS RRB Clerk Previous Papers<\/a><\/p>\n Question 7:\u00a0<\/b>There are 20 compartments in a train. If each compartment can contain upto maximum of 30 people and each compartment should have different number of people except 2 compartments and no compartment should have less than 12 people in it. If the maximum number of passengers are to be filled in the train then what is the average number of people in each compartment ?<\/p>\n a)\u00a021.45<\/p>\n b)\u00a021<\/p>\n c)\u00a020<\/p>\n d)\u00a020.45<\/p>\n e)\u00a022.45<\/p>\n Question 8:\u00a0<\/b>Average runs scored by Sehwag in his first 85 innings was 52 and in each of the next 5 innings till 100 innings his average gradually increased as 53 and then to 54 and then to 55 at the end of 100 innings. What is the difference between the runs scored by him in 86 to 90 innings and the last 10 innings ?<\/p>\n a)\u00a0340<\/p>\n b)\u00a0350<\/p>\n c)\u00a0360<\/p>\n d)\u00a0370<\/p>\n e)\u00a0380<\/p>\n Question 9:\u00a0<\/b>Average of four numbers is 147. First number is (5\/17) times the second number and the second number is 36 less than the third number and the third number is 52 more than the half of the fourth number. What is the LCM of four numbers ?<\/p>\n a)\u00a0175480<\/p>\n b)\u00a0175000<\/p>\n c)\u00a0174480<\/p>\n d)\u00a0175420<\/p>\n e)\u00a0175440<\/p>\n Question 10:\u00a0<\/b>There are 4 people in a family whose ages are P,P+12,P+36 and P+60 years and after \u2018y\u2019 years the oldest of the family leaves the family and the family has only 3 people but the average of the family doesn\u2019t change. After another \u2018y\u2019 years the average of the age of 3 people is 47 years then what is the value of P ?<\/p>\n a)\u00a09<\/p>\n b)\u00a010<\/p>\n c)\u00a011<\/p>\n d)\u00a012<\/p>\n e)\u00a013<\/p>\n Quantitative Aptitude formulas PDF<\/a><\/p>\n 520 Banking Mocks – Just Rs. 499<\/a><\/p>\n Question 11:\u00a0<\/b>A circle is inscribed inside a rectangle of length 12 cm and breadth Y cm. The circle touches the inner part of the breadth of the rectangle. If the difference between their areas is $\\dfrac{320}{7}$ sq.cm, then find Y.<\/p>\n a)\u00a06 cm<\/p>\n b)\u00a08 cm<\/p>\n c)\u00a04.75 cm<\/p>\n d)\u00a07.84 cm<\/p>\n e)\u00a09 cm<\/p>\n Question 12:\u00a0<\/b>The area of a rectangle is 68 sq.cm. If the length of the rectangle is reduced by 3 cm and the breadth of the rectangle is increased by 5 cm, then its area reduces to 12 less than half of its original area. If its length is increased by 2 cm and its breadth is reduced by 4 cm, then its area will be 10 more than its original area. Find its length.<\/p>\n a)\u00a07.65 cm<\/p>\n b)\u00a03.42 cm<\/p>\n c)\u00a04 cm<\/p>\n d)\u00a06 cm<\/p>\n e)\u00a0None of these<\/p>\n Question 13:\u00a0<\/b>If two sides of a triangle are 6 cm and 4 cm then how many integral values are possible for the third side ?<\/p>\n a)\u00a06<\/p>\n b)\u00a07<\/p>\n c)\u00a08<\/p>\n d)\u00a09<\/p>\n e)\u00a010<\/p>\n Question 14:\u00a0<\/b>A person standing on the top of a building of height 20m observes the angle of elevation of the top of other building as 60 degrees and angle of depression of the bottom the same building as 30 degrees. What is the difference between the heights of two building ?<\/p>\n a)\u00a010m<\/p>\n b)\u00a040m<\/p>\n c)\u00a050m<\/p>\n d)\u00a030m<\/p>\n e)\u00a060m<\/p>\n Question 15:\u00a0<\/b>A cuboid has its length, breadth and height in the ratio of 4:5:6 If the length is decreased b 25% ,breadth is increased by 20% and height is increased by 33.33 % then what is the percentage change in total surface area of the cuboid ?<\/p>\n a)\u00a021.1%<\/p>\n b)\u00a023.8%<\/p>\n c)\u00a022.3%<\/p>\n d)\u00a021.6%<\/p>\n e)\u00a022.6%<\/p>\n Question 16:\u00a0<\/b>Ram invested in stocks in three investments in the ratio 6:8:15 and he got 20% profit on 1st investment and 37.5% loss in second investment and 10% profit in the third investment. Find his overall profit or loss percentage.<\/p>\n a)\u00a020.36%<\/p>\n b)\u00a019.25%<\/p>\n c)\u00a021.78%<\/p>\n d)\u00a019.65%<\/p>\n e)\u00a021.37%<\/p>\n Question 17:\u00a0<\/b>A man has 2 sons and the ages of them are 12 years and 15 years.He deposits total of Rs 3,60,000 in the bank for a time period that is until any one of them reaches 18 years and the money is deposited in the ratio of their present ages.What is the difference between the total amount they receive after the stipulated time if rate of interest is 5% simple interest ?<\/p>\n a)\u00a042000<\/p>\n b)\u00a046000<\/p>\n c)\u00a040000<\/p>\n d)\u00a032000<\/p>\n e)\u00a036000<\/p>\n Question 18:\u00a0<\/b>A person purchases 200 apples at Rs 300 per dozen.He sold 40% of the apples at Rs 30 per each apple and 25% of the remaining apples were spoiled and in the remaining apples half were sold at Rs 40 per each and the other half were sold at 35 per each. Find the profit\/loss percent<\/p>\n a)\u00a015.5%<\/p>\n b)\u00a014%<\/p>\n c)\u00a015.7%<\/p>\n d)\u00a014.8%<\/p>\n e)\u00a015%<\/p>\n Question 19:\u00a0<\/b>A trader got 10%,20% and 30% profit by selling three article A,B and C respectively.If the sum of cost price by A and B is twice that of C and the difference between the cost price of A and B(A>B) is 40% of the cost price of A. What is the overall profit of the trader ?<\/p>\n a)\u00a060.2%<\/p>\n b)\u00a060.8%<\/p>\n c)\u00a060.4%<\/p>\n d)\u00a060.6%<\/p>\n e)\u00a070%<\/p>\n Question 20:\u00a0<\/b>To meet the demand during summer, a milkman mixed water with pure milk in 2:5. Then he sold the same mixture at 20% more than normal price. By what percent his revenue will increase during summers? (Water is available without any cost)<\/p>\n a)\u00a060 percent<\/p>\n b)\u00a050 percent<\/p>\n c)\u00a040 percent<\/p>\n d)\u00a068 percent<\/p>\n e)\u00a063 percent<\/p>\n 18000 Free Solved Questions – Banking Study Material<\/a><\/p>\n 35 IBPS RRB PO Mocks @ Rs. 149. Enroll Now<\/a><\/p>\n Answers & Solutions:<\/strong><\/span><\/p>\n 1)\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 2)\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 3)\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 4)\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 5)\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 6)\u00a0Answer\u00a0(C)<\/strong><\/p>\n $1798.98\\div8.01\\times51.965\\div779.97-390.001\\div25.87 \\simeq 1800\\div8\\times52\\div780-390\\div26$<\/p>\n $= \\Large\\frac{1800\\times52}{8\\times780}-\\frac{390}{26}$<\/p>\n $= 15-15 = 0$<\/p>\n 7)\u00a0Answer\u00a0(A)<\/strong><\/p>\n It is given that in each compartment maximum of 30 people can be accommodated and only two compartments can have the same number of people and maximum number of people have to be accommodated in the train and so 30 people should be in two compartments and in the other compartments we should have 29,28,27\u2026…12 8)\u00a0Answer\u00a0(E)<\/strong><\/p>\n Given average for first 85 innings is 52 9)\u00a0Answer\u00a0(E)<\/strong><\/p>\n let the four numbers be a,b,c and d 10)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Present average=(P+P+12+P+36+P+60)\/4 11)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Given, Length of the rectangle = 12 cm Area of the circle = $\\dfrac{22}{7}\\times\\dfrac{Y}{2}\\times\\dfrac{Y}{2} = \\dfrac{11Y^2}{14} sq.cm$<\/p>\n Given, $12Y – \\dfrac{11Y^2}{14} = \\dfrac{320}{7}$<\/p>\n => $168Y – 11Y^2 = 640$ 12)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Let the original length of the rectangle be l cm and the original breadth of the rectangle be b cm. If the length is increased by 2 and breadth is reduced by 4, Therefore, Original length of the rectangle = 4 cm.<\/p>\n 13)\u00a0Answer\u00a0(B)<\/strong><\/p>\n For a triangle to form the sum of two sides should be greater than the third side and the difference of two sides should be less than the third side 14)\u00a0Answer\u00a0(E)<\/strong><\/p>\n Let the height of other building be \u2018h\u2019 15)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Let the length,breadth and height of the cuboid be 4x,5x and 6x respectively 16)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Let the three investments be 6x,8x and 15x. 17)\u00a0Answer\u00a0(B)<\/strong><\/p>\n From the given information the time period is 18-15=3 years 18)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Given CP per dozen=300 19)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Let the cost price of A,B and C be a,b and c 20)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Let us assume that on a normal day he sells milk at x rupee per litre and he sells y litre of milk He mixed water with milk in ratio of 2:5 so net quantity = $\\frac{2+5}{5} * y$ = 1.4y So revenue generated in summers = 1.4y * 1.2x = 1.68xy
\n$17+143+235+38 = 433$<\/p>\n
\n$77+5 = 82$<\/p>\n
\n$=$ $22500-9$ $=$ $22491$<\/p>\n
\nSo the total number of passengers=30+30+29+….12
\n=30+(30(31)\/2)-(11(12)\/2)
\n=30+465-66
\n=399+30
\n=429
\nTotal number of compartments=20
\nAverage=429\/20
\n=21.45<\/p>\n
\nSum of the runs scored by Sehwag for first 85 innings=85*52
\n=4420
\nSum of the runs scored by him in first 90 innings =90*53
\n=4770
\nRuns scored by him from 86 to 90 innings=4770-4420
\n=350
\nRuns scored by him in first 100 innings=100*55
\n=5500
\nRuns scored by him in last 10 innings=5500-4770
\n=730
\nRequired difference=730-350
\n=380<\/p>\n
\nGiven a+b+c+d=147*4
\na+b+c+d=588
\na=5b\/17
\nb=c-36
\nc=b+36
\nc=(d\/2)+52
\nb+36=(d+104)\/2
\n2b+72=d+104
\nd=2b-32
\n(5b\/17)+b+b+36+2b-32=588
\n(5b\/17)+4b+4=588
\n((5b+68b)\/17) =584
\n73b=584*17
\nb=136
\na=40
\nc=172
\nd=240
\nLCM of 40,136,172,240 is 175440<\/p>\n
\n=(4P+108)\/4
\nAfter \u2018y\u2019 years P+60 age person will leave so now average becomes
\n((P+y+P+12+y+P+36+y)\/3)=(4P+108)\/4
\n(3P+3y+48)\/3 =P+27
\nP+y+16=P+27
\ny=27-16
\ny=11
\nNow (3P+6y+48)\/3 =47
\n3P+6y+48=141
\n3P=141-48-66
\n3P=27
\nP=9<\/p>\n![](\"https:\/\/cracku.in\/media\/uploads\/MEn.JPG\")
<\/figure>\n
\nBreadth of the rectangle = Y cm
\nArea of the rectangle = 12Y sq.cm
\nThe circle touches the inner part of the breadth of the rectangle.
\nThen, Diameter of the circle = Breadth of the rectangle
\nRadius of the circle = $\\dfrac{Y}{2}$ cm<\/p>\n
\n=> $11Y^2-168Y+640 = 0$
\nSolving above equation, we get,
\nY = 8 or Y = 7.27.
\nTherefore, The breadth of the rectangle can be either 8 cm or 7.27 cm.
\nHence, From the options given, Breadth of the rectangle = 8 cm<\/p>\n
\nThen its area = lb = 68 sq,cm
\nIf the length is reduced by 3 cm and breadth is increased by 5 cm,
\n(l-3)(b+5) = 68\/2-13 = 22
\nlb+5l-3b-15 = 22
\n68+5l-3b-15 = 22
\n5l-3b+53 = 22
\n5l-3b+31 = 0 — (1)<\/p>\n
\n(l+2)(b-4) = 68+10 = 78 sq.cm
\nlb-4l+2b-8 = 78
\n-4l+2b = 18
\n4l-2b+18 = 0 — (2)
\nSolving (1) and (2)
\n2l – 8 = 0
\nl = 4<\/p>\n
\nTherefore (6+4)>s>(6-4)
\n10>s>2
\nSo all the values between 2 and 10 are possible as third side
\ni.e 3,4,5,6,7,8 and 9
\nTherefore 7 values are possible<\/p>\n
\nDistance between the buildings be \u2018d\u2019
\nFrom the figure we have $\\tan60$=d\/20
\nd=$\\sqrt{3}20$
\nFrom the figure we have in the upper triangle $\\tan60$=x\/d
\n$\\sqrt{3}$=x\/$\\sqrt{3}20$
\nx=60 cm
\nTherefore h=60+20
\nh=80 cm
\nDifference=80-20
\n=60 cm<\/p>\n
\nTotal surface area of the cube=2lb+2bh+2lh
\n=$2*(20*x^{2})+2*(24*x^{2})+2*(30*x^{2})$
\n=$(40+48+60)x^{2}$
\n=$148x^{2}$
\nNew length of cuboid=4x*3\/4
\n=3x
\nNew breadth of cuboid=5x*(120\/100)
\n=6x
\nNew height of cuboid=6x*(4\/3)
\n=8x
\nTotal surface area of cuboid with new dimensions=$2*(18*x^{2})+2*(48*x^{2})+2*(24*x^{2})$
\n=$(36+96+48)x^{2}$
\n=$180x^{2}$
\nPercentage change in the total surface area=((180-148)\/148)*100
\n=800\/37
\n=21.6%<\/p>\n
\nGiven on 1st investment profit=20%
\nHis selling price=6x*1.2
\n=7.2x
\nGiven on 2st investment loss=37.5%
\nHis selling price=8x*11\/8
\n=11x
\nGiven on 3rd investment profit=10%
\nHis selling price=15x*1.1
\n=16.5x
\nTotal CP=29x
\nTotal SP=7.2x+11x+16.5x
\n=34.7x
\nProfit percent=((34.7x-29x)\/29x)*100
\n=19.65%<\/p>\n
\nSo the money is distributed in the ratio 4:5 i.e On 1st son it is (4\/9)*360000=160000
\nOn the second son it is (5\/9)*360000=200000
\nInterest for 1st son=160000*3*5\/100
\n=24000
\nInterest for 2n son=200000*3*5\/100
\n=30000
\nAmount received by 1st son=160000+24000
\n=184000
\nAmount received by 2nd son=200000+30000
\n=230000
\nDifference=230000-184000
\n=46000<\/p>\n
\nFor each apple cost=300\/12
\n=Rs 25
\nTotal cost price of apples=200*25
\n=5000
\nHe sold 40% of apples at Rs 30 per each
\nSo (40\/100)*200 =80 apples
\n80*30=Rs 2400
\n25% of remaining apples are spoiled i.e (1\/4)*120=30 apples
\nSo total leftover apples=120-30
\n=90
\nSo 45 apples are sold at 40 per each and the other 45 apples at 35 per each
\nTherefore total SP=45*40+45*35
\n=1800+1575
\n=Rs 3375
\nTotal SP=3375+2400
\n=5775
\nProfit percent=((5775-5000)\/(5000))*100
\n=775\/50
\n=15.5 %<\/p>\n
\nSP of A=1.1a
\nSP of B=1.2b
\nSP of C=1.3c
\nTotal SP=1.1a+1.2b+1.3c
\nTotal CP=a+b+c
\nAso given a+b=2c
\na-b=0.4a
\nb=0.6a
\na+b=2c
\n1.6a=2c
\nc=0.8a
\nTotal SP=1.1a+1.2(0.6a)+1.3(0.8a)
\n=1.1a+0.72a+1.04a
\n=3.86a
\nTotal CP=a+0.6a+0.8a
\n=2.4 a
\nTotal profit percent=((3.86a-2.4a)\/(2.4a))*100
\n=60.8%<\/p>\n
\nSo revenue generated on normal day = x*y = xy<\/p>\n
\nNew selling price during summer = 1.2x rupees per litre<\/p>\n
\nHence we can say that revenue will go up by 68 percent.<\/p>\n