1.A number divided by 68 gives the
quotient 269 and remainder zero. If the same number is divided by 67, the
remainder is :
(1) 0
(2) 1
(3) 2
(4) 3
2.If the sum and difference of two
numbers are 20 and 8 respectively, then the difference of their squares is:
(1) 12
(2) 28
(3) 80
(4) 160
3.If ⋆ is an operating such that x ⋆ y = 3x + 2y, then 2 ⋆
3 + 3 ⋆ 4 is equal to :
(1) 18
(2) 29
(3) 32
(4) 38
4.In a school, the average age of
students is 6 years, and the average age of 12 teachers is 40 years. If the
average age of the combined group of all the teachers and students is 7 years,
then the number of students is :
(1) 396
(2) 400
(3) 408
(4) 416
5.The average age of 30 boys in a
class is 15 years. One boy aged 20 years, left the class, but two new boys came
in his place whose ages differ by 5 years. If the average age of all the boys
now in the class still remains 15 years, the age of the younger newcomer is :
(1) 20 years
(2) 15 years
(3) 10 years
(4) 8 years
6.Given that 10% of A’s income = 15%
of B’s income = 20% of C’s income. If sum of their income is Rs. 7800. Then B’s
income is :
(1) Rs. 3600
(2) Rs. 3000
(3) Rs. 2400
(4) Rs. 1800
7.If two numbers are respectively
20% and 50% of a third number, what is the ratio between the two numbers
?
(1) 5 : 2
(2) 2 : 5
(3) 1 : 5
(4) 1 : 2
8.If a number x is 10% less than
another number y and y is 10% more than 125, then x is equal to
(1) 150
(2) 143
(3) 140.55
(4) 123.75
9.Two numbers are in the ratio 2 :
3. If 2 is subtracted from the first and 2 is added to the second, the ratio
becomes 1 : 2. The sum of the numbers is :
(1) 30
(2) 28
(3) 24
(4) 10
10.The ratio of incomes of two
persons is 5 : 3 and that of their expenditures is 9 : 5. If they save Rs, 2600
and Rs. 1800 respectively, their incomes are :
(1) Rs. 8000; Rs. 4800
(2) Rs. 6000; Rs. 3600
(3) Rs. 10000; Rs. 6000
(4) Rs. 9000; Rs. 5400
Answers and Solution:
1. (2)
Number = 269*68
= 269 * (67+1)
= 269 * 67 + 269
clearly,remainder is obtained on dividing 269 by 67 that is 1.
2. (4)
Let x+y = 20 and x-y = 8
S0 (x+y)(x-y) = 20*8
=> x^2-y^2 = 160
3. (2)
Here x*y = 3x+2y
so 2*3+3*4 = (3*2+2*3)+(3*3+2*4)
= 12+17=29
4. (1)
Let the number of students be x.then,
=> 7 = (x*6+12*40)/(x+12)
=7x + 84 = 6x + 480
=> x = 396
5. (2)
6. (3)
10% of A = 15% 0f B = 20* 0f C
=> 10A = 15B = 20C
=> 10A/60 = 15B/60 = 20C/60
=> A/6 = B/4 = C/3
So A : B : C = 6 : 4 : 3
so 6x + 4x + 3x = 7800
=> 13x = 7800
x = 600
B's income = 4x = 600*4 = 2400
7.(2)
8. (4)
9. (1)
10. (1)
Let the incomes of two persons be Rs.5x and 3x rupess and their expenditure be Rs.9x and 5x respectively,
As given
5x - 9y = 2600.......(1)
3x - 5y = 1800.......(2)
by solve
then, x = 1600
So First person's income = 1600*5 = 8000
Second's person income = 1600*3 = 4800
1. (2)
Number = 269*68
= 269 * (67+1)
= 269 * 67 + 269
clearly,remainder is obtained on dividing 269 by 67 that is 1.
2. (4)
Let x+y = 20 and x-y = 8
S0 (x+y)(x-y) = 20*8
=> x^2-y^2 = 160
3. (2)
Here x*y = 3x+2y
so 2*3+3*4 = (3*2+2*3)+(3*3+2*4)
= 12+17=29
4. (1)
Let the number of students be x.then,
=> 7 = (x*6+12*40)/(x+12)
=7x + 84 = 6x + 480
=> x = 396
5. (2)
6. (3)
10% of A = 15% 0f B = 20* 0f C
=> 10A = 15B = 20C
=> 10A/60 = 15B/60 = 20C/60
=> A/6 = B/4 = C/3
So A : B : C = 6 : 4 : 3
so 6x + 4x + 3x = 7800
=> 13x = 7800
x = 600
B's income = 4x = 600*4 = 2400
7.(2)
8. (4)
9. (1)
10. (1)
Let the incomes of two persons be Rs.5x and 3x rupess and their expenditure be Rs.9x and 5x respectively,
As given
5x - 9y = 2600.......(1)
3x - 5y = 1800.......(2)
by solve
then, x = 1600
So First person's income = 1600*5 = 8000
Second's person income = 1600*3 = 4800
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