quants

Number System Question types

1.What is the remainder when the sum 1! + 2! + 3! + …+ 100! is divided by 40?

2.Find the number of zeros at the end of 22 x 35 x 625 x 20 x 8 x 160.

3.BODMAS operation

4.Express 0.9(63) in the form of a fraction.  Conversion into fractions

5.In how many ways can 30030 be written as a product of two factors?

6.Find the number of factors of 3025.

solution--3025 =  (5)(605)= (5)(5)(121) = 52 112

Number of factors = (2+1)(2+1) = 9

7.Base conversion concept

8.HCF LCM OF fraction types

9.Number of zeros at the end

Example Find the number of zeros at the end of 22 x 35 x 625 x 20 x 8 x 160.

               To find the number of zeros at the end of a number, find out the powers of 2 and 5 in the expression.

As 10 = 2 × 5, so the minimum power of 2 or 5 will give us the number of zeroes at the end of the final number.

⇒ 22 x 35 x 625 x 20 x 8 x 160.= (2 x 11) x (5 x 7) x 54 x (22 x 5) x 23 x (25 x 5)

= 211 x 57 x 11 x 7

Here minimum power of 2 or 5 is 7, hence, there will be 7 zeros at the end of the expression.

TIME AND SPEED

1.Trains and Distance

model 1. starts with simple problem , finding distance , speed and time

HINTS: 1.see of conversions carefully m/s---->km/hr(use 18/5)  and km/hr---->m/s (use 5/18)

2.move in same direction then relative speed is the difference , if opposite direction then the speed is summation of the speed

model 2. using relative speeds

HINTS:3. in such relative problems when the speed is in varied format and distance and time in other format, conversion shall be done for sum or difference of the speed.

eg- train 125 long and man walking in same direction with 5km/hr speed .cross in 10 seconds. find the length of train 

                125 + 0                                              x=45 km/hr  (we have considered the

             -----------------   = 10            --------------->                        conversion for whole thing

                (5+x)5/18                                                                so only the x value in km/hr)

4. Always check for the end of question for the form(km, m, or s, min, hrs) of answer the at the end of question

model 3.trains starting at different times and and speeds given find the time they meet?

model 4. Time taken for two trains going from A to B and B to A given.find when they meet?

model 5. If two trains (or bodies) start at the same time from points A and B towards each other and after crossing they take a and b sec in reaching B and A respectively, then:

(A's speed) : (B's speed) = (sq.rt(b) :sq.rt(a))

model 6: Finding average of list of given speed-- eg.10 ,20, 30, 60 kmph..Find the average speed


model 7 finding ratios of speed or time ---->use Speed=1/time
model 8- for the same distance the average speed is the formula 2xy/(x+y)

area problems

type-1 

A rectangular park 60 m long and 40 m wide has two concrete crossroads running in the middle of the park and rest of the park has been used as a lawn. If the area of the lawn is 2109 sq. m, then what is the width of the road?

A. 2.91 m B 3 m C. 5.82 m D. None of these Answer: Option B Explanation: Area of the park = (60 x 40) m2 = 2400 m2. Area of the lawn = 2109 m2.  Area of the crossroads = (2400 - 2109) m2 = 291 m2. Let the width of the road be x metres. Then, 60x + 40x - x2 = 291 (remember this formulation)  x2 - 100x + 291 = 0  (x - 97)(x - 3) = 0  x = 3. 2. The diagonal of a rectangle is 41 cm and its area is 20 sq. cm. The perimeter of the rectangle must be: A. 9 cm B. 18 cm C. 20 cm D. 41 cm Answer: Option B Explanation: l2 + b2 = 41. Also, lb = 20. (l + b)2 = (l2 + b2) + 2lb = 41 + 40 = 81(use this way ..do not go for l=20/b and keep doing shit)  (l + b) = 9.  Perimeter = 2(l + b) = 18 cm. 3. A towel, when bleached, was found to have lost 20% of its length and 10% of its breadth. The percentage of decrease in area is: percentage change in area $=(-20-10+\frac{20\times 10}{100})\%=-28\%$ i.e., area is decreased by $28\%$ $\mathrm{<br>}$ 4. What is the least number of squares tiles required to pave the floor of a room 15 m 17 cm long and 9 m 2 cm broad? A. 814 B. 820 C. 840 D. 844 Answer: Option A Explanation: Length of largest tile = H.C.F. of 1517 cm and 902 cm = 41 cm. Area of each tile = (41 x 41) cm2.  Required number of tiles = 1517 x 902 = 814. 41 x 41