Practice Quantitative aptitude - quantitative aptitude Online Quiz (set-2) For All Competitive Exams
Q-1) A sum of money lent out at simple interest amounts to Rs.720 after 2 years and to Rs.1020 after a further period of 5 years. The sum is :
(a)
(b)
(c)
(d)
Principal + SI for 2 years = Rs.720 .... (i)
Principal + SI for 7 years = Rs.1020 .....(ii)
Subtracting equation (i) from (ii) get,
SI for 5 years
= Rs.(1020 - 720) = Rs.300
SI for 2 years
= Rs.300 × $2/5$ = Rs.120
Principal = Rs.(720 - 120) = Rs.600
Using Rule 12,
P = $({A_2T_1 - A_1T_2}/{T_1 - T_2})$
= $({1020 × 2 - 720 × 7}/{2 - 7})$
= $({2040 - 5040}/{- 5})$
= ${- 3000}/{- 5}$ = Rs.600
Q-2) The least number which when divided by 5, 6, 7 and 8 leaves a remainder 3, but when divided by 9 leaves no remainder is
(a)
(b)
(c)
(d)
The LCM of 5, 6, 7 and 8 = 840
∴ Required number = 840 k + 3 which is exactly divisible by 9 for some value of k.
Now, 840 k + 3 = 93 × 9 k + (3k + 3)
When k = 2, 3k + 3 = 9, which is divisible by 9.
∴ Required number = 840 × 2 + 3 = 1683
Q-3) Two numbers are in the ratio of 3 : 5. If 9 be subtracted from each, then they are in the ratio of 12 : 23. Find the numbers.
(a)
(b)
(c)
(d)
According to the question,
${3x - 9}/{5x - 9} = 12/23$
(Numbers = 3x and 5x )
69x - 207 = 60x - 108
9x = 207 - 108 = 99
x = 11
Required numbers
3 × 11 = 33 and 5 × 11 = 55
Using Rule 35,
Here, a = 3, b = 5, c = 12,
d = 23, x = 9
1st Number = ${xa(d-c)}/{ad-bc}$
= ${9 × 3(23 - 12)}/{3 × 23 - 5 × 12}$
= ${27 × 11}/{69 - 60}$
= ${27 × 11}/9$= 33
2nd Number= ${xb(d-c)}/{ad-bc}$
= ${9 × 5(23 - 12)}/{3 × 23 - 5 × 12}$
= ${45 × 11}/{69 - 60}$
= ${45 × 11}/9$ = 55
Numbers are 33, 55
Q-4) A man and his wife appear for an interview for two posts. The probability of the husband's selection is $1/7$ and that of the wife's selection is $1/5$ . The probability that only one of them will be selected is
(a)
(b)
(c)
(d)
Probability that only husband is selected
= P(H) P($\ov{W}$) = $1/7(1 - {1/5}) = 1/7 × 4/5 = 4/{35}$
Probability that only wife is selected
= P($\ov{H}$) P(W) = $(1 - {1/7})(1/5) = 6/7 × 1/5 = 6/{35}$
∴ Probability that only one of them is selected
=$4/{35} + 6/{35} = {10}/{35} = 2/7$
Q-5) Profit earned by an organisation is distributed among officers and clerks in the ratio of 5 : 3. If the number of officers is 45 and the number of clerks is 80 and the amount received by each officer is Rs.25,000, what was the total amount of profit earned?
(a)
(b)
(c)
(d)
(e)
Amount received by all the officers
= 45 × 25000 = 11,25,000
Amount received by each clerk
= $3/5 × 25000$ = 15000
Amount received by all the clerks
= 80 × 15000 = 12,00,000
Total amount of profit earned
= 11,25,000 + 12,00,000 = Rs.23.25 lakh.
Q-6) Gautam goes to office at a speed of 12 kmph and returns home at 10 kmph. His average speed is :
(a)
(b)
(c)
(d)
Here distances are same.
∴ Average speed = $({2xy}/{x + y})$ kmph
= $({2 × 12 × 10}/{12 + 10})$ kmph
= $(240/22)$ kmph = 10.9 kmph
Q-7) Two alloys are both made up of copper and tin. The ratio of copper and tin in the first alloy is 1 : 3 and in the second alloy is 2 : 5. In what ratio should the two alloys be mixed to obtain a new alloy in which the ratio of tin and copper be 8 : 3 ?
(a)
(b)
(c)
(d)
By rule of alligation
Required ratio
= $1/77 : 1/44$ = 4 7
Q-8) Which is the largest among the numbers $√5 , 3√7 , 4√13$
(a)
(b)
(c)
(d)
$√5 , 3√7 , 4√13$
$√5$
$3√7 =√{9×7}=√63$
$√^4{13}=√{4×4×13}=√{208}$
Clearly,$√5<3√7<4√13$
Q-9) If 4 litres of water is evaporated on boiling from 12 litres of salt solution containing 7 percentage salt, the percentage of salt in the remaining solution is
(a)
(b)
(c)
(d)
In 12 litres salt solution,
Salt = ${7 × 12}/100 = 0.84$ units
Water = ${93 × 12}/100 = 11.16$ units
After evaporation,
Percentage of salt = $0.84/8 × 100 = 10.5$%
Q-10) A man sold a book at a profit of 10%. If he had charged Rs.45 more, his profit percentage would have been 25%. Find the C.P. of the book.
(a)
(b)
(c)
(d)
(e)
15% of x = 45
$x = {45 × 100}/15$ = 300
Q-11) If f(x) is a polynomial with all coefficients are integers and constant term 10 having a factor (x - k), where k is an integer, then what is the possible value of k?
(a)
(b)
(c)
(d)
Given that, f(x) is a polynomial with constant term 10 and all coefficient are integer. Let $k, k_1, k_2, ...... k_{n - 1}$ be roots of nth degree polynomial.
Now (x - k) is a factor of f(x), where k is an integer.
Product of roots = ${\text"Constant term"}/{\text"Coefficient of" x^n}$
i.e., $k. k_1. k_2......k_{n - 1} = {10}/1$ Coefficient of $x^n$ is 1
⇒ $k. k_1. k_2....k_{n -1}$ = 10 = 5.2.1....1.
Therefore, the possible value of k is 5.
Q-12) Average of first five prime numbers is
(a)
(b)
(c)
(d)
Required average
= ${2+3+5+7+11}/5$
= $28/5$ = 5.6
Q-13) If $√2$ = 1.4142… is given, then the value of $7/{(3 +√2)}$ correct upto two decimal places is
(a)
(b)
(c)
(d)
$7/{(3 +√2)}={7(3-√2)}/{(3+√2)(3-√2)}$
[Rationalising the denominator]
=${7(3-√2)}/{9-2}$
$[(a+b)(a-b)=a^2-b^2]$
=$3-√2$
= 3 - 1.4142 = 1.5858
= 1.59 (correct to two decimal places)
Q-14) 100 oranges are bought for Rs.350 and sold at the rate of Rs.48 per dozen. The percentage of profit or loss is :
(a)
(b)
(c)
(d)
C.P. of 100 oranges = Rs.350
S.P. of 12 oranges = Rs.48
S.P. of 100 oranges
= $48/12 × 100$ = Rs.400
Profit = Rs. (400 - 350) = Rs.50
Profit % = $50/350 × 100 = 100/7 = 14{2}/7%$
Using Rule 13,
Here, a = 100, x = 350, b = 100, y = $48/12$ × 100 = 400
Gain% = $({ay - bx}/{bx})$ × 100%
= ${100 × 400 - 100 × 350}/{100 × 350}$ × 100%
= ${40 - 35}/35 × 100%$
= $100/7% = 14{2}/7$%
Q-15) The speed of a car is 54 km/hr. What is its speed in m/sec?
(a)
(b)
(c)
(d)
1 kmph = $5/18$ m/sec
54 kmph = $5/18 × 54$
= 15 m/sec.
Q-16) Two places P and Q are 162 km apart. A train leaves P for Q and simultaneously another train leaves Q for P. They meet at the end of 6 hours. If the former train travels 8 km/hour faster than the other, then speed of train from Q is
(a)
(b)
(c)
(d)
Speed of train starting from Q = x kmph
Speed of train starting from P = (x + 8) kmph
According to the question,
PR + RQ = PQ
(x + 8) × 6 + x × 6 = 162
[Distance = Speed × Time]
6x + 48 + 6x = 162
12x = 162 - 48 = 114
x = $114/12 = 19/2 = 9{1}/2$ kmph
Q-17) A thief is noticed by a policeman from a distance of 200m. The thief starts running and the policeman chases him. The thief and the policeman run at the rate of 10 km./ hr and 11 km./hr respectively. What is the distance between them after 6 minutes ?
(a)
(b)
(c)
(d)
Using Rule 12,If a train of length l m passes a bridge/platform of 'x' m in $t_1$ sec, then the time taken by the same train to cross another bridge/platform of length 'y' m is,Time taken = $({l + y}/{l + x})t_1$
Relative speed of police
= 11 – 10 = 1 kmph
= $5/18$ m/sec
Distance decreased in 6 minutes
= $5/18$ × 6 × 60 = 100 m
Distance remained between them
= 200–100 = 100 m
Q-18) The difference between the simple interest received from two different sources on Rs.1500 for 3 years is Rs.13.50. The difference between their rates of interest is:
(a)
(b)
(c)
(d)
Let $r_1$, and $r_2$ be the required rate of interest
Then, ${13.50} = {1500 × 3 × r_1}/100$
– ${1500 × 3 × r_2}/100 = 4500/100(r_1 - r_2)$
$r_1 - r_2 = 135/450 = 27/90$
= $3/10$ = 0.3%
Using Rule 13,
$P_1 = Rs.1500, R_1 , T_1$ = 3 years.
$P_2 = Rs.1500, R_2 , T_2$ = 3 years.
S.I. = Rs.13.50
13.50 = ${1500 × R_2 × 3 - 1500 × R_1 × 3}/100$
$1350/100 = {4500(R_2 - R_1)}/100$
$R_2 - R_1 = 1350/4500 = 27/90$
= $3/10$ = 0.3%
Q-19) Out of the three given numbers, the first number is twice the second and thrice the third. If the average of the three numbers is 121, what is the difference between the first and the third number?
(a)
(b)
(c)
(d)
(e)
Let the third number be = x
∴ First number = 3x and
second number = ${3x}/2$
According to the question.
or, $3x + {3x}/2 + x = 3 × 121$
or, ${6x + 3x + 2x}/2$ = 3 × 121
or, ${11x}/2$ = 3 × 121
∴ $x = {3 × 121 × 2}/11$ = 66
∴ Third number = 66
Required difference
= 3x – x = 2x = 2 × 66 = 132
Q-20) A boy has a few coins of denominations 50 paise, 25 paise and 10 paise in the ratio 1 : 2 : 3. If the total amount of the coins is 6.50, the number of 10 paise coins is
(a)
(b)
(c)
(d)
Ratio of the value of coins
= $1/2 : 2/4 : 3/10$ = 5 : 5 : 3
Value of the 10-paise coins
= Rs.$(3/13 × 6.50)$ = Rs.1.5
Number of 10-paise coins
= 1.5 × 10 = 15
Q-21) A boy saves Rs.4.65 daily. What is the least number of days in which he will be able to save an exact number of rupees ?
(a)
(b)
(c)
(d)
Through option 4.65 when multiplied by = 20 gives whole number
Q-22) A boat takes 19 hours for travelling downstream from point A to point B and coming back to point C, mid way between A and B. If the velocity of the stream is 4 km/hr and the speed of the boat in still water is 14 km/hr. then the distance between A & B is
(a)
(b)
(c)
(d)
Downstream speed = 14 + 4 = 18 km/hr
Upstream speed = 14 – 4 = 10 km/hr
Let the distance between A and B = 'x' km
∴ $x/18+{x/2}/10=19$
∴ $x/18+x/20=19$
${10x+9x}/180=19$
${19x}/180=19$ ⇒ x=180km
Q-23) The average of 5 consecutive even numbers A, B,C,D and E is 52. What is the product of B and E?
(a)
(b)
(c)
(d)
(e)
Let the five consecutive even numbers be
x, x + 2, x + 4, x + 6 and x + 8 respectively.
According to the question,
x + x + 2 + x + 4 + x + 6 + x + 8 = 5 × 52
or 5x + 20 = 260
or 5x = 260 – 20
or $x = 240/5$ = 48
∴ B = x + 2 = 48 + 2 = 50
and E = x + 8 = 48 + 8 = 56
∴ B × E = 50 × 56 = 2800
Q-24) The average marks in English subject of a class of 24 students is 56. If the marks of three students were misread as 44, 45 and 61 of the actual marks 48, 59 and 67 respectively, then what would be the correct average?
(a)
(b)
(c)
(d)
(e)
Correct average
= ${(24 × 56) + (48 + 59 + 67) - (44 + 45 + 61)}/24$
= ${1344 + 174 - 150}/24 = 1368/24$ = 57
Q-25) An empty fuel tank of a car was filled with A type petrol. When the tank was half-empty, it was filled with B type petrol. Again when the tank was half-empty, it was filled with A type petrol. When the tank was half-empty again, it was filled with B type petrol. What is the percentage of A type petrol at present in the tank ?
(a)
(b)
(c)
(d)
Let the capacity of the tank be 100 litres.
Then, Initially : A type petrol = 100 litres.
After first operation :
A type petrol = $(100/2)$ = 50 litres;
B type petrol = 50 litres.
After second operation :
A type petrol = $(50/2 + 50) = 75$litres;
B type petrol = (50/2) = 25 litres
After third operation :
A type petrol = $(75/2)$ = 37.5 liters;
B type petrol = $(25/2 + 50)$ = 62.5 litres.
∴ Required percentage = 37.5%.
Q-26) In a city, 40% of the people are illiterate and 60% are poor. Among the rich, 10% are illiterate. The percentage of the illiterate poor population is
(a)
(b)
(c)
(d)
Let the population of the city be 100.
Total illiterate people = 40
Poor people = 60; Rich people = 40
Illiterate rich people = ${40 × 10}/100$ = 4
Illiterate poor people = 40 – 4 = 36
Required percent = $36/60 × 100$ = 60%
Q-27) If 5 men or 7 women can earn Rs.5,250 per day, how much would 7 men and 13 women earn per day ?
(a)
(b)
(c)
(d)
5 men ≡ 7 women
[Both earn same amount in 1 day]
7 men ≡ $7/5 × 7 = 49/5$ women
7 men + 13 women
= $49/5 + 13 = 114/5$ women
Now, Since, 7 women ≡ Rs.5250
$114/5$ women ≡ $5250/7 × 114/5$ = Rs.17100
Q-28) A is faster than B. A and B each walk 24 km. The sum of their speeds is 7 km/hr and the sum of times taken by them is 14 hours. Then A's speed is equal to:
(a)
(b)
(c)
(d)
Let, A's speed = x kmph.
B's speed = (7 - x) kmph
Time = $\text"Distance"/ \text"Speed"$
According to the question,
$24/x + 24/{7 - x} = 14$
$24({7 - x + x}/{x(7 - x)})$ = 14
${24 × 7}/{x(7 - x)}$ = 14
x (7 - x) = 12 = 4 × 3 or 3 × 4
x (7 - x) = 4 (7 - 4) or 3 (7 - 3)
x = 4 or 3
A's speed = 4 kmph.
Q-29) If one-third of one-fourth of a number is 15, then three-tenth of the number is
(a)
(b)
(c)
(d)
Let the number be x.
= $x/{3 × 4}$ = 15
x = 15 × 3 × 4 = 180
Now, required number
= ${3/10}x = 3/10$ × 180 = 54
Q-30) In a two–digit number, the digit at the unit’s place is 1 less than twice the digit at the ten’s place. If the digits at unit’s and ten’s place are interchanged, the difference between the new and the original number is less than the original number by 20. The original number is
(a)
(b)
(c)
(d)
Ten’s digit = x
Unit’s digit = 2x – 1
Original number = 10x + (2x – 1) = 12x – 1
New number = 10 (2x – 1) + x
= 20x – 10 + x = 21x – 10
(21x – 10) – (12x + 1) = 12x – 1 – 20
9x – 9 = 12x – 21
3x = 12 ⇒ x = 4
Original number = 12x – 1 = 12 × 4 – 1 = 47