SAT (Stage-1) Mock Test

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SAT (Stage-1) Mock Test - 70

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Weekly Quiz Competition

Question 1
1 / -0

Find the value of

.

A
2/3

B
3/2

C
5/6

D
4/5

Solution
Question 2
1 / -0

If x = 2 + √3 and y = 2 – √3, then the value of 1/x + 1/y is

A
3

B
4

C
2√3

D
2

Solution
Question 3
1 / -0

Find the largest number among 4^1/4, , and √2.

A
4^1/4

B

C
√2

D
All are equal.

Solution
Question 4
1 / -0

The salary of B is 10% more than the salary of A, the salary of C is 10% more than the salary of B, and the salary of D is 10% more than the salary of C. The total salary of A and D is Rs. 2100 more than that of B and C. What is the salary of B?

A
Rs. 1,00,000

B
Rs. 1,10,000

C
Rs. 1,11,000

D
Rs. 1,01,000

Solution

Let salary of A be x.
Then:
Salary of B = 1.1x
Salary of C = 1.21x
Salary of D = 1.331x
A + D = 2.331x
B + C = 2.31x
According to the question:
(A + D) - (B + C) = 2100
0.021x = 2100
x = 1,00,000
Therefore, salary of B is Rs. 1,10,000.
Question 5
1 / -0

What is the area of the shaded portion in the given figure?

A
196 cm²

B
224 cm²

C
240 cm²

D
254 cm²

Solution

Area of shaded region EH = Area of quadrant AEH - Area of △ AEH

Area of the required shaded region = 4 × Area of shaded region EH

224 cm²
Question 6
1 / -0

Find the area of the shaded region in the following figure.

A
80 cm²

B
85 cm²

C
70 cm²

D
75 cm²

Solution

Area of the shaded region = Area of rectangle ABCD - Area of rectangle EFGH
Area of the shaded region = ((10 × 8) - (1 × 5)) cm²
= (80 - 5) cm² = 75 cm²
Question 7
1 / -0

On dividing x⁴ + 2x² - 3x + 7 by a polynomial g(x), the quotient and the remainder were x² + 5 and -3x + 22, respectively. The polynomial g(x) was

A
x - 3

B
x² - 3

C
x² + 3

D
x + 3

Solution

x⁴ + 2x² - 3x + 7 = (x² + 5)g(x) - 3x + 22
Let g(x) = x² + ax + b
Thus, x⁴ + 2x² - 3x + 7 = (x² + 5)(x² + ax + b) - 3x + 22
Or, x⁴+ 2x² - 3x + 7 = x⁴ + ax³+ bx² + 5x² + 5ax + 5b - 3x + 22
Or, x⁴ + 2x² - 3x + 7 = x⁴ + ax³ + x²(b + 5) + x(5a - 3) + 5b + 22
On comparing coefficients, we get
a = 0, b + 5 = 2 or b = -3
Thus, g(x) = x² - 3
Question 8
1 / -0

A tradesman marks an article at a price which would give him a profit of 20% on the cost price. To the favoured customers, he makes a deduction of 5% from the marked price. What actual profit does he receive from the sale of an article to a favoured customer for which the latter pays him $28.50?

A
$3.50

B
$5.70

C
$5.00

D
$5.75

Solution

Let the CP of the article be $x.
MP = x + 20% of x = $1.2x
When a discount is given on this MP,
SP of the article = 95/100× 1.2x = $1.14x
But 1.14x = 28.50
So, x = 25
CP of the article = $25
SP = $28.50
Profit = $3.50
Question 9
1 / -0

What profit will be made by selling an article at a certain price, if there is a loss of 10% incurred by selling it at 2/3 of that price?

A
10%

B
15%

C
25%

D
35%

Solution

Let SP of the article be 3s.

Let CP of the article be c.

SP when a loss of 10% is incurred = 2/3 × 3s = 2s
Question 10
1 / -0

The ratio of apples to mangoes is 5 : 9. If 35 mangoes are sold, then the ratio becomes 10 : 11. Find the number of apples.

A
50

B
25

C
40

D
85

Solution

According to the given condition,
Number of apples is 5x and that of mangoes is 9x.
Then, from the question:

⇒ 11x = 18x - 70
⇒ 7x = 70
⇒ x = 10
Hence, number of apples = 5x = 50
Question 11
1 / -0

Find the rate percent of the simple interest per annum if the sum borrowed becomes double in 5 years.

A
15%

B
16%

C
20%

D
24%

Solution

Let the principal amount be 'P' and rate be r.
Amount after 5 years = 2P
Simple interest = A - P = 2P - P = P
According to the question,

r = 20%
Rate = 20%
Question 12
1 / -0

The slant height of a cone is increased by k%. If radius remains the same, the curved surface area will be increased by

A
k%

B
2k%

C
k/2 %

D
Remains same

Solution

Curved surface area of cone = π × r × l
As slant height is directly proportional to the curved surface area being radius constant, the curved surface area will be increased by k%.
Question 13
1 / -0

The radius of a sphere is increased by k%. Its surface area increases by

A
k%

B
k²%

C

D

Solution
Question 14
1 / -0

A man can row 18 km/hr in still water. The time taken by him to go upstream is twice the time taken by him to come downstream. Find the speed of the stream.

A
6.5 km/hr

B
2 km/hr

C
5 km/hr

D
6 km/hr

Solution

Let the speed of the stream be x km/hr.
Let the distance travelled be 'd' .
Speed in upstream = (18 – x) km/hr
Speed in downstream = (18 + x) km/hr
According to question,
Time in upstream = 2(Time in downstream)

18 + x = 2(18 – x)
18 + x = 36 – 2x
3x = 18
x = 6 km/hr
Speed of the stream = 6 km/hr
Question 15
1 / -0

The sum of the first n terms, the first term and the common difference of an AP are 418, 13 and 5, respectively. What is the value of n?

A
11

B
12

C
13

D
17

Solution

First term a = 13
Let the number of terms be 'n'.
Common difference, d = 5
Sum = 418

836 = n(26 + 5n - 5)
836 = 5n² + 21n
5n² + 21n - 836 = 0
5n² + 76n - 55n - 836 = 0
n(5n + 76) - 11(5n + 76) = 0
(n - 11)(5n + 76) = 0
⇒n = 11 or -76/5
Since, n cannot be negative and a fraction,
Hence, n = 11