Test:Measurement - 1 - CTET & State TET MCQ

Money left = 61.50 - 42.75 = 18.75 Rs

Length of smaller pieces = 11×0.45 = 4.95
Hence,

ribbon left with Reeta =17.85 − 4.95 = 12.90

To solve this problem, we need to calculate the maximum possible weight of the parcel that can be sent using foreign mail with a postage of Rs. 136.

The postage rates for foreign mail are:

• First 60 g: Rs. 32.50
• Every additional 50 g or part thereof: Rs. 4.50

Step 1: Calculate the remaining postage after the first 60 g.

Given that the first 60 g costs Rs. 32.50, subtract this from the total postage:

136 − 32.50 = 103.50 Rs.

Step 2: Determine how many additional 50 g units can be covered by Rs. 103.50.

Each additional 50 g costs Rs. 4.50, so divide the remaining postage by this amount:

Step 3: Calculate the total weight.

Add the weight for the first 60 g and the additional weight:

60 g+ (23×50 g) = 60 g +1150 g=1210 g

So, the maximum possible weight of the parcel is 1210 g.

Soni can cycle from point A to point B and return to point A in 10 minutes.
so for 1 round with cycle it would take 5 minutes
 She can cycle and walk back in 18 minutes.
it would take 18 - 5 minutes to walk , so 13 minutes for walk
It will take 26 mins for her to walk there and walk back.

Cost of 3 pieces = 345

It takes 30 minutes to walk one kilometer.

weight of hazelnut = 4.75kg

weight of raisins = 0.75 kg

total weight = 4.75 +0.75 = 5.5kg

No of box = 5

therefore weight of each box = total weight / no of box

= 5.5kg/ 5

= 1.1kg

Weight of Aman's bag = 4.72−2.14 = 2.58

on left side 12 balls
on right side 7 balls and kg
so as both side weigth are equal so equating them
12x=7x+40
5x=40
x=8
The mass of 20 such balls will be 160kg

The instrument used to measure Blood Pressure is Sphygmomanometer.

  Let capacity of beaker=x Then, 

  Saving per kg = 195/13 = 15

Length of 5 smaller pieces = 
∴ cloth left with tailor = meters

28 m is larger becuase 1m = 1000 cm

Thus 28 m = 2800 cm which is greater than 2750cm

Balls required will be 6

5 for balancing the triangle and one for the other. 

=15600sec.

390g would be split in 60g and 250g 
and 250g = 5 x 50g
so for first 60g 12.5Rs and for every 50g x 2.5 
12.50 + 5 × 2.5 = Rs.25

1 bucket = 6 jugs = 3 litres. 

So. 4 buckets will be  = 3*4 = 12 litres. 

One box = 4 buckets. 

Therefore, answer is 12 litres. 

Perimeter of the quadrilateral = a + b + c + d
Perimeter of the quadrilateral = 4.6 + 8.4 + 7.3 + 4.7 = 25

Let quantity of water in tank B be x litres.
Then quantity of water in the tank A=6x litres
Total water in both the tanks =x+6x
=140 liters
⇒7x=140⇒x=20
∴ Water in tank B=20 litres
⇒ Water in tank A=20×6=120 litres
So, quantity of water to be transferred from tank A to tank B, so that each tank contains 70 litres= 70 −20=50 litres
Therefore, 50 liters  of water should be transferred.