a) 30°
b) 75°
c) Both 30° and 75°
d) Either 30° or 75°
⇒ tan 46° tan 29° tan2 A tan 61° tan 44° = tan A
⇒ tan 46° tan 29° tan2 A tan (90 - 29)° tan (90 - 46)° = tan A
⇒ tan 46° tan 29° tan2 A cot 29° cot 46° = tan A
⇒ tan 46° tan 29° tan2 A(1/tan 29°) (1/tan 46°) = tan A
⇒ tan2 A = tan A
⇒ tan2 A - tan A = 0
⇒ tan A (tan A - 1) = 0
⇒ tan A = 0 and tan A = 1
⇒ A = 0° and 45°
⇒ tan (A + B) = (√3 + 1)/(√3 - 1)
⇒ tan (A + B) = tan 75° [∵ tan75 = tan(45 + 30) = (√3 + 1)/(√3 - 1)]
⇒ (A + B) = 75°
When A = 0°:
⇒ B = 75°
When A = 45°
⇒ B = 30°2. If a + b + c = 45 and ab + bc + ca = 254. Find the value of a^2 + b^2 + c^2 = ?
a) 1452
b) 1517
c) 1652
d) 1256
3. The circumferences of two circles are touching externally. The distance between their centres is 12 cm. The radius of one circle is 7 cm. Find the diameter (in cm) of the other circle.
Source: safalta
a) 8
b) 10
c) 12
d) 6
When two circles touch each other externally, the distance between their centres is equal to the sum of their radiuses,
⇒ Distance = 12 = 7 + Radius of other circle
⇒ Radius of other circle = 12 – 7 = 5
∴ Diameter of other circle = 2(5) = 10 cm
4. The speed of a train relative to a moving car is 70km/hr more. The car accelerates and starts moving at a speed 10% more than before. Both now travel for 3 hours and the train covers a total distance of 750 km. How much distance did the car travel?
a) 564km
b) 574km
c) 584km
d) 594km
The train covers 750 km in three hours.
The speed of train = 750/3 = 250 km/hr
Initially the train had a relative speed of 70km/hr to the car.
Initial speed of car = 250 – 70 = 180 km/hr
The speed of car now is increased by 10%.
New speed of car = 180 + 18 = 198km/hr
At this speed the car travels for 3 hrs.
∴ The distance travelled by car = 198 × 3 = 594km
5. Find the greatest number that will divide 390, 495 and 300 without leaving a remainder.
a) 5
b) 10
c) 15
d) 20
Lets factorize each one them
390 = 2 × 3 × 5 × 13
300 = 2 × 2 × 3 × 5 × 5
495 = 3 × 3 × 5 × 11
Hence we can see only two numbers common two all three are 3 × 5 = 15
∴ 15 is the greatest number that divides 390, 495 and 300 without leaving a remainder.
6. The monthly salaries of A and B are in the ratio 8 : 9. If they get salary increment of Rs. 4000 each, the ratio of new monthly salaries of A and B becomes 10 : 11. What is the new monthly salary of A?
a) Rs. 30000
b) Rs. 15000
c) Rs. 20000
d) Rs. 25000
⇒ Let the monthly salaries of A and B be 8x and 9x respectively
⇒ after increment new salary of A and B be (8x + 4000) and (9x + 4000) respectively
⇒ According to question
⇒ (8x + 4000) / (9x + 4000) = 10 / 11
⇒ 11 ⨯ (8x + 4000) = 10 ⨯ (9x + 4000)
⇒ 88x + 44000 = 90x + 40000
⇒ 2x = 4000
⇒ x = Rs. 2000
∴ New salary of A = 8× 2000 + 4000 = Rs. 20000
7. 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 places of both digits are interchanged, the new number formed is 27 more than the original number. The original number is
a) 47
b) 37
c) 45
d) 55
Let the ten’s digit and unit digit of the two-digit number be x and y respectively.
⇒ Two digit number = 10x + y
According to the given information,
y = 2x – 1
⇒ 2x – y = 1 …..(i)
If the places of the digits are interchanged,
10y + x = 10x + y + 27
⇒ 9y - 9x = 27
⇒ y – x = 3 …..(ii)
We get from 1 and 2,
y = 7
x = 4
∴ The number is 47.
8. Of three positive numbers, the ratio of first and second is 4 ∶ 7, that of second and third is 2 ∶ 3. The product of first and third is 4200. What is the sum of the three numbers?
a) 225
b) 205
c) 215
d) 235
Let the three numbers be x, y and z.
Ratio of x and y = 4 ∶ 7
Ratio of y and z = 2 ∶ 3
Then, x ∶ y ∶ z = 8 ∶ 14 ∶21
Let the three number be 8a, 14a and 21a.
The product of x and z = 4200
⇒ x × z = 4200
⇒ 8a × 21a = 4200
⇒ a2 = 25
⇒ a = 5
First number = x = 8a = 40
Second number = y = 14a = 70
Third number = z = 21a = 105
The sum of three numbers = 40 + 70 + 105 = 215
9. The average of 11 numbers is 8. If a number 12 is removed, then what will be the new average?a) 7.2
b) 7.4
c) 7.6
d) 7.8
Average of 11 numbers = 8
∴ Sum of 11 numbers = 11 × 8 = 88
After a number 12 is removed
Sum of remaining 10 numbers = 88 - 12 = 76
∴ Average of remaining 10 numbers
= 76/10
= 7.6
10. The speed of the boat in still water is 18 km/hr and the speed of the current is 2 km/hr. If A man rows his boat upstream for 3 hours and downstream for 2 hours. Find the total distance travelled by him?
a) 82 km
b) 84 km
c) 86 km
d) 88 km
Let speed of boat in still water is SB and speed of current is SC and Speed of boat in upstream and downstream is SU and SD.
Speed of boat in still water SB = 18 km/hr
Speed of current, SC = 2 km/hr
Speed of boat in upstream, SU = 18 - 2 = 16 km/hr
Speed of boat in downstream, SD = 18 + 2 = 20 km/hr
∴ Total distance travelled = 3 × 16 + 2 × 20 = 48 + 40 = 88 km