# APTITUDE TEST 2

## About the topic Aptitude Test 2

In this online Aptitude Test 2,we give the questions whose qualities  are in high standard and practicing these questions will definitely make the students to reach their goals.

Example Quiz 1

1. Find the number of prime factors of 610 X 717 X 5527.
 (A) 90                   (B) 91 (C) 92                   (D) 93

2. Two trains running at 60 kmph and 48 kmph cross each other in 15 seconds when they run in opposite direction. When they run in the same direction, a person in the faster train observes that he crossed the slower train in 36 seconds. Find the length of the two trains (in meters).
 (A) 300, 130              (B) 310, 115 (C) 320, 110              (D) 330, 120

3. A & B can do a work in 15 days, B & C in 30 days and A & C in 18 days. They work together for 9 days and then A left. In how many more days, can B and C finish the remaining work ?
 (A) 9 days             (B) 8 days (C) 7 days             (D) 6 days

4. A man traveled from the village to the post office at the rate of 25 k mph and walked back at the rate of 4 kmph. If the entire journey had taken 5 hours 48 minutes, find the required distance of the post office from the village.
 (A) 20 km             (B) 21km (C) 22 km             (D) 23 km

5. The present age of a father is 3 years more than three times the age of his son. Three years hence, father's age will be 10 years more than twice the age of the son. Find the present age of the father.
 (A) 35 yrs                (B) 34 yrs (C) 33 yrs                (D) 32 yrs

6. The average of four consecutive even numbers is 27. Find the largest of these numbers.
 (A) 28                     (B) 30 (C) 32                     (D) 34

7. A boat takes 19 hours for traveling downstream from point A to point B and coming back to a point C midway between A and B. If the velocity of the stream is 4 kmph and the speed of the boat in still water is 14 kmph, what is the distance between A and B ?
 (A) 150 km                 (B) 160 km (C) 170 km                 (D) 180 km

8. John weighs 56.7 kilograms. If he is going to reduce his weight in the ratio 7:6, find his new weight.
 (A) 48.6 kg                (B) 49.6 kg (C) 50.6 kg                (D) 51.6 kg

9. Find the ratio in which, water to be mixed with milk to gain 20% by selling the mixture at cost price.
 (A) 1:5                          (B) 5:1 (C) 1:7                          (D) 1:7

10. 15% of income of A is equal to 25% of income of B and 10% of income of B is equal to 30% of income of C. If income of C is \$ 1600, then total income of A, B and C is
 (A) \$ 14200              (B) \$ 14300 (C) \$ 14400              (D) \$ 14500

Explanation

 Question No.1 Question No.2 Question No.3 Question No.4 Question No.5 Question No.6 Question No.7 Question No.8 Question No.9 Question No.10 jQuery UI Accordion - Default functionality .ui-widget-overlay,.ui-state-disabled,ui-button{background:#fff;border:1px solid #fff;color:#b9cd6d;font-weight:bold} .ui-widget-overlay,.ui-state-disabled,ui-button{background:#fff;border:1px solid #fff;color:#b9cd6d;font-weight:bold} From 610 × 717 × 5527, we have to write each base in terms of multiplication of its prime factors. That is,=(2X3)10×(7)17×(5X11)27 =210X310X717X527X1127 The no. of prime factors = sum of the exponents = 10+10+17+27+27 = 91 Hence, the number of prime factors is 91. jQuery UI Accordion - Default functionality .ui-widget-overlay,.ui-state-disabled,ui-button{background:#fff;border:1px solid #fff;color:#b9cd6d;font-weight:bold} .ui-widget-overlay,.ui-state-disabled,ui-button{background:#fff;border:1px solid #fff;color:#b9cd6d;font-weight:bold} When the two train running in opposite direction, relative speed = 60+48 = 108 kmph = 108X5/18 m/sec = 30 m/sec Sum of the lengths of the two trains = sum of the distances covered by the two trains in the above relative speed Sum of the lengths of the two trains = 30X15 = 450 m When the two trains running in the same direction, relative speed = 60-48 =12 kmph = 12X5/18 = 10/3 m/sec When the two trains running in the same direction, a person in the faster train observes that he crossed the slower train in 36 seconds. The distance he covered in 36 seconds in the relative speed is equal to the length of the slower train. Length of the slower train = 36X10/3 = 120 m Length of the faster train = 450-120 = 330 m Hence, the length of the two trains are 330m and 120m jQuery UI Accordion - Default functionality .ui-widget-overlay,.ui-state-disabled,ui-button{background:#fff;border:1px solid #fff;color:#b9cd6d;font-weight:bold} .ui-widget-overlay,.ui-state-disabled,ui-button{background:#fff;border:1px solid #fff;color:#b9cd6d;font-weight:bold} We can apply L.C.M method to solve this problem Total work = 90 units (L.C.M of 15,30,18,9) (A+B) can complete 6 units/day (90/15 = 6) (B+C) can complete 3 units/day (90/30 = 3) (A+C) can complete 5 units/day (90/18 = 5) By adding, we get 2(A+B+C) = 14 units/day (A+B+C) = 14/2 = 7 units/day work done by A = (A+B+C)-(B+C)=7-3=4 units/day work done by B = (A+B+C)-(A+C)=7-5=2 units/day work done by C = (A+B+C)-(A+B)=7-6=1 unit/day work done by (A+B+C)in 9 days = 9X7 = 63 units. Balance work = 90-63 = 27 units This 27 units of work to be completed B & C. Because A left after 9 days of work. No. of days taken by B & C to complete the balance 27 units of work = 27/3 = 9 days. (Because B&C can complete 3 units of work per day) jQuery UI Accordion - Default functionality .ui-widget-overlay,.ui-state-disabled,ui-button{background:#fff;border:1px solid #fff;color:#b9cd6d;font-weight:bold} .ui-widget-overlay,.ui-state-disabled,ui-button{background:#fff;border:1px solid #fff;color:#b9cd6d;font-weight:bold} Average speed = 2pq/(p+q), here p=25 q=4 = (2X25X4)/(25+4) Therefore, average speed = 200/29 km/hr And 5 hour 48 min = 548⁄60hrs = 29/5 hours Distance = Speed X Time Distance = (200/29)X(29/5) = 40 km Distance covered in 5 hrs 48 min = 40 km Hence,distance of the post office from the village = 40/2 = 20km jQuery UI Accordion - Default functionality .ui-widget-overlay,.ui-state-disabled,ui-button{background:#fff;border:1px solid #fff;color:#b9cd6d;font-weight:bold} .ui-widget-overlay,.ui-state-disabled,ui-button{background:#fff;border:1px solid #fff;color:#b9cd6d;font-weight:bold} Let "x" be the present age of the son.Then the present age of father is (3x+3) 3 years hence, father's age will be 10 years more than twice the age of the son (3x+3)+3 = 2(x+3)+10 3x+6 = 2x+16 x = 10 To find the present age of the father, plug x = 10 in (3x+3) Present age of the father = 3(10)+3 = 33 years jQuery UI Accordion - Default functionality .ui-widget-overlay,.ui-state-disabled,ui-button{background:#fff;border:1px solid #fff;color:#b9cd6d;font-weight:bold} .ui-widget-overlay,.ui-state-disabled,ui-button{background:#fff;border:1px solid #fff;color:#b9cd6d;font-weight:bold} If "x' be the first even number, then the four consecutive even numbers are x, x+2, x+4, x+6 Average of the four consecutive even numbers = 27 (x+x+2+x+4+x+6)/4 = 27 (4x+12) = 108 4x = 96 ===> x = 24 Hence the largest even number = x+6 = 30 jQuery UI Accordion - Default functionality .ui-widget-overlay,.ui-state-disabled,ui-button{background:#fff;border:1px solid #fff;color:#b9cd6d;font-weight:bold} .ui-widget-overlay,.ui-state-disabled,ui-button{background:#fff;border:1px solid #fff;color:#b9cd6d;font-weight:bold} From the given information, we have Speed downstream = (14+4) = 18 kmph Speed upstream = (14-4) = 10 kmph Let "x" be the distance between A and B x/18 + (x/2)/10 =19 (Hint: Time = Distance/Speed) x/18 + x/20 =19 (10x + 9x)/180 = 19 (L.C.M of 18,20 is 180) 19x/180 = 19 x = 180 Hence the distance between A and B is 180 km. jQuery UI Accordion - Default functionality .ui-widget-overlay,.ui-state-disabled,ui-button{background:#fff;border:1px solid #fff;color:#b9cd6d;font-weight:bold} .ui-widget-overlay,.ui-state-disabled,ui-button{background:#fff;border:1px solid #fff;color:#b9cd6d;font-weight:bold} Original weight of John = 56.7 kg (given) He is going to reduce his weight in the ratio 7:6 [Hint:If a quantity increases or decreases in the ratio a:b, then new quantity = 'b' of the original quantity divided by 'a'] His new weight = (6x56.7)/7 = 6x8.1 = 48.6 kg. jQuery UI Accordion - Default functionality .ui-widget-overlay,.ui-state-disabled,ui-button{background:#fff;border:1px solid #fff;color:#b9cd6d;font-weight:bold} .ui-widget-overlay,.ui-state-disabled,ui-button{background:#fff;border:1px solid #fff;color:#b9cd6d;font-weight:bold} Let the cost price of 1 ltr pure milk be \$1 Now we take some quantity of milk (less than 1 ltr),add some water and make it to be 1 ltr mixture Let "x" be the money we invest for milk in 1 ltr of milk-water mixture Since the gain is 20%, selling price = x + 20%of x Selling price = 120% of x = (120/100)x = (6/5)x But the mixture is sold at the cost price of pure milk So, we have (6/5)x = 1 x = 5/6 Therefore cost price of the milk in the mixture = \$(5/6 Cost price of the water = \$0 Rule to find the ratio for producing mixture = (d-m):(m-c) (d-m):(m-c) = 1-5/6:5/6-0 = 1/6:5/6 = 1:5 So, water and milk to be mixed in the ratio to gain 20% is 1:5 jQuery UI Accordion - Default functionality .ui-widget-overlay,.ui-state-disabled,ui-button{background:#fff;border:1px solid #fff;color:#b9cd6d;font-weight:bold} .ui-widget-overlay,.ui-state-disabled,ui-button{background:#fff;border:1px solid #fff;color:#b9cd6d;font-weight:bold} Let A,B and C be the incomes of A,B and C respectively From the given information, we have C = \$1600 10% of B = 30% of C (10/100)B = (30/100)X1600 B = \$ 4800 15% of A = 25% of B (15/100)A = (25/100)X4800 A = \$8000 A+B+C = 8000+4800+1600 = 14400 Hence, the total income of A,B and C is \$14400

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