Complete the following table using given information. - Vidyadip

Question

Complete the following table using given information.

✓

Answer

(i) MV=10+7=₹ 17 (ii) at discount of 25- 16=₹ 9 (iii) FV= ₹ 5.

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

More "Complete the following activities and rewrite it : (2M)" from P-1 Financial Planning→P-1 Financial Planning — all question types→Maths STD 10 question papers→Maharashtra Board STD 10 question papers→Maharashtra Board question paper generator→

Similar questions

Complete the following activity to solve the simultaneous equations

seg PQ || seg DE, A(Δ PQF) = 20 units, PF = 2 DP, then find A(DPQE) by completing the following activity.
A(Δ PQF) = 20 units, PF = 2 DP, Let us assume DP = x. ∴ PF = 2x

In Δ FDE and Δ FPQ,
∠FDE ≅ ∠ .......... corresponding angles
∠FED ≅ ∠ ......... corresponding angles
∴ Δ FDE ~ Δ FPQ .......... AA test

→

Complete the following activity to find the 27 th term or the A.P. 9. 4. -1,-6,-11, ........

Complete the following activity to find the value of x.

Write the values of the following trigonometric ratios.
$\tan 45^{\circ}=\frac{⬜}{⬜}$

Write the correct number in the given boxes from the following A. P.
1, 8, 15, 22, . . .
$\text { Here } a =\square, t _1=\square, t _2=\square, t _3=\square, $
$ t _2- t _1=\square-\square=\square $
$ t _3- t _2=\square-\square=\square \therefore d =\square$

In figure 1.66, seg PQ || seg DE, A(Δ PQF) = 20 units, PF = 2 DP, then find A(DPQE) by completing the following activity.
A(Δ PQF) = 20 units, PF = 2 DP, Let us assume DP = x. ∴ PF = 2x
$DF = DP +\square=\square+\square=3 x$
In Δ FDE and Δ FPQ,
∠FDE ≅ ∠ .......... corresponding angles
∠FED ≅ ∠ ......... corresponding angles
∴ Δ FDE ~ Δ FPQ .......... AA test
$\therefore \frac{ A (\Delta FDE )}{ A (\Delta FPQ )}=\frac{\square}{\square}=\frac{(3 x )^2}{(2 x )^2}=\frac{9}{4}$
$A (\Delta FDE )=\frac{9}{4} A(\Delta FPQ )=\frac{9}{4} \times \square=\square$
$A (\square DPQE )= A (\Delta FDE )- A (\Delta FPQ )$
$\quad=\square-\square$
$\quad=\square$

Fill in the gaps and complete.
If α, β are roots of quadratic equation,

→

In ΔPQR seg PM is a median. Angle bisectors of ∠PMQ and ∠PMR intersect side PQ and side PR in points X and Y respectively. Prove that XY || QR.

Complete the proof by filling in the boxes. In ΔPMQ, ray MX is bisector of ∠PMQ.

.......... (I) theorem of angle bisector.
In ΔPMR, ray MY is bisector of ∠PMR.

.......... (II) theorem of angle bisector.
But $\frac{M P}{M Q}=\frac{M P}{M R}$ .......... M is the midpoint QR, hence MQ = MR.
$\therefore \frac{ PX }{ XQ }=\frac{ PY }{ YR }$
∴ XY || QR .......... converse of basic proportionality theorem.

→

In ∆ABC,∠ACB is obtuse angle, seg AD ⊥ seg BC. Prove that:
AB² = BC² + AC² - 2BC×CD