ChetanaSforum  # careerskk

Members

7

## Community Reputation

0 Neutral

• Rank
Newbie
1. Find two C Aptitude questions (solved) from previous Accenture placement papers1) Which of the following statements is false?a)Pointers are designed for storing memory addressesb)Arrays are passed by value to functionsc)Both of the above are falseAnswer: Both of the above are false2) Which of the following statements is true regarding static variables?a) Which of the following statements is true regarding static variables? Only local variables can be defined as static variablesc) Scope and life of global variables is limited to the function to which they are declaredAnswer: Both local and global variables can be defined as static variables
2. ## Sample Campus Placement Questions Solved from Quantitative Aptitude

Sorry in the above questio â€“ refers to difference.
3. ## Sample Campus Placement Questions Solved from Quantitative Aptitude

Two sample questions on model of HCL and TCS placement papers have been given. The questions are from Discounts chapter.1. The bankers gain on a certain sum due 1 Â½ year hence is 3/25 of the bankers discount .The rate percent isSol:- Let B.D = 1 then B.G = 3/25 T.D = (B.D â€“ B.G) = (1 â€“ 3/25) = 22/25 sum = (1 * 22/25) / (1 â€“ 22/25) = 22/3 S.I on Rs 22/3 for 1 Â½ year is 1. Rate = (100 * 1) / (22/3 * 3/2) = 9 1/9%2. The bankers gain of a certain sum due 2 years hence at 10% per annum is Rs 24 .The percent worth isSol:- T.D = (B.G * 100) / (Rate * Time) 00) / (10 * 2) = 120. P.W = (100 *T.D) / (Rate * Time) = (100 * 120) /(10 * 2) = 600
4. Below question is from section on Discounts from Quantitative aptitude Question: The Present worth of a bill due something hence is Rs 1110 and the true discount on the bill is Rs.110 . Find the bankes discount & the bankers gain. Answer: T.D = sqrt (P.W * B.G) B.G = (T.D)^2 /P.W = (110 * 110) / 1110 = 11 B.D = (T.D + B.G) = (110 + 11) = Rs 121
5. Below are three questions compiled on the format of TCS placement papers.1) In a 100 race,A can give B 10 mand C 28 m.In the same race Bcan give C?Sol: A:B =100 :90 A :C=100 :72 B:C =B/A *A/C =90/100*100/72 =90/72 When B runs 90 C runs 72 when B runs 100 C runs =72*100/90 B beats C by 20m2) In a 100m race ,A beats B by 10 m and C by 13 m.In the race of 180m. B will beat c by? Sol :A : B =100 :90 A/B =100/90 A/C =100/87 B/C =B/A *A/C =90 /87 When B runs 90m C runs 87 When B runs 180 m then C runs =87 *180/90 B beats C by 180-174= 6m3) In a race of 200 m, A can beat B by 31 m and C by 18m.In arace of 350 m, C will beat B by?Sol : In a race of 200 m A :B =200:169 A :C =200 :182 C/B =C/A*A/B =182/200*200/169 When C runs 182 m B runs 169 when C runs 350 m B runs =350*169/182 =325m
6. Wipro, TCS and HCL placement papers contain quantitative aptitude questions. 1) In a game of 100 points A can give B 20 points and C 28 points then B can give C? Sol: In a game of 100 points.A :B =100 :80 A :C =100 :72 B/C=B/A*A/C =80/100*100/72 = 80/72 when B runs 80m C runs 72 when B runs 100m C runs =100*72/80 =90 B can give C 10 points in agame of 100. 2)At a game of billiards,A can give B 15 points in 60 and A can give C 20 points in60.How any points can B give C in a game of 90? Sol: A:B =60:45 A:C =60:40 B/C =B/A*A/C =45/60*60/40 =90/80 B can give C 10 points in a game of 90. 3) In a game of 80 points,A can give B 5 points and C 15 points. Then how many points B can give C in a game of 60? Sol: A :B =80 :75 A :C =80:65 B/C=B/A*A/C = 75/80*80/65 = 15/13 B:C =60:52
7. Below I have given few quantitative aptitude questions (solved) from ratios and proportions. Such questions can be expected on Wipro Placement Papers. 1. In what ratio must rice at Rs 9.30 per Kg be mixed with rice at Rs 10.80 per Kg so that the mixture be worth Rs 10 per Kg? Solution: C.P of 1 Kg rice of 1st kind 930 p C.P of 1 Kg rice of 2n d kind 1080p Mean Price 1000p 80 70 Required ratio=80:70 = 8:7 2. How much water must be added to 60 liters of milk at 11/2 liters for Rs 20 so as to have a mixture worth Rs 10 2/3 a liter? Solution: C.P of 1 lit of milk = 20*2/3 = 40/3 C.P of 1 lit of water 0 C.P of 1 lit of milk 40/3 Mean Price 32/3 8/3 32/3 Ratio of water and milk =8/3 : 32/3 = 1:4 Quantity of water to be added to 60 lit of milk =1/4*60=15 liters. 3. In what ratio must water to be mixed with milk to gain 20% by selling the mixture at cost price? Solution:Let the C.P of milk be Re 1 per liter Then S.P of 1 liter of mixture = Re.1 Gain obtained =20%. Therefore C.P of 1 liter mixture = Rs(100/120*1) =5/6 C.P of 1 liter of water 0 C.P of 1 liter of milk1 Mean Price 5/6 1/6 5/6 Ratio of water and milk =1/6 : 5/6 = 1:5. 4. In what ratio must a grocer mix two varieties of pulses costing Rs 15 and Rs 20 per Kg respectively so as to get a mixture worth Rs 16.50 per Kg? Solution: Cost of 1 Kg pulses of 1 kind 15 Cost of 1 Kg pulses of 2nd kind 20 Mean Price Rs 16.50 3.50 1.50 Required ratio =3.50 : 1.50 = 35:15 = 7:3.
×